Math and Logic Puzzles

[Who]

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-------------------------------
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[Who]

I'm trying to see if stuff happens as the numbers get bigger. There appeared to be more 0s in a row in the 900s and 9000s than in the >100s, so when in doubt, add more 0s:
10^19+1
10^19+2
10^19+3
10^19+4
10^19+5
10^19+6
10^19+7
10^19+8
10^19+9
10^19+10
10^19+11
10^19+12
10^19+13
10^19+14
10^19+15
10^19+16
10^19+17
10^19+18
10^19+19
10^19+20

[Who]

Aha, it takes longer with bigger numbers, that's probably more info than just yes or no.

[Who]

The thing is, at an early age, I was taught with the clear implication that a whole number and a decimal number could not be equated without rounding. If I learned this stuff when I was 5 years old, we probably wouldn't be having this argument.

You expect the stuff you learn at an early age to be true? When I was 5, I was told that negative numbers did not exist. I was also told that Santa Claus did exist. I have since learned that 99% of what I learned at age 5 was BS.

The nice thing about math, is that it is not about what you are told, it is about what can be proven. You have to assume a few things to start with, but once you assume those you can find everything else.

[Who]

@Kage
Fun problem! Took me a while.

Spoiler: My solution

A, Would you say "Ja" if I asked you if B was random?
If A says "Ja", either A or B is random, and ask C the next question. If A says "Da", either A or C is random, and ask B the next question.
Would you say "Ja" if I asked you if A was random?
If the answerer says "Ja", A is random, otherwise the other one is random.
Then ask one of the non-random ones "Would you say 'ja' if I asked you if you were false?". If they say "Ja", they are false, if they say "Da", they are true.

[Who]

An interesting follow-up to the blue-eyes puzzle is what exact piece of information is given by the elder that allows them to leave. (Surely everyone else, seeing at least 49 blue eyed people, already knows that there is at least 1 blue eyed person.)

The elder gives them a starting time.

[Who]

Also it means that everyone knows this, and that there is at least one person with blue eyes.

[Who]

Do you include rectangles with an area of zero as rectangles?

Spoiler: If not

A rectangle is defined by two diagonal points, giving ((m+1)(n+1))*((m+1)(n+1)-n-m+1)/4 for the number of rectangles without wrap-around. If the grid is assumed to wrap around, there is another variable: Which border, if any, of the grid it crosses. For any set of two points, there could be 4 options (Vertically, horizontally, both, and stays inside the grid). Thus, the final number is ((m+1)(n+1))*((m+1)(n+1)-n-m+1)

Spoiler: if so

A rectangle is defined by two diagonal points, giving ((m+1)(n+1))*((m+1)(n+1)-n-m+1)/4 for the number of rectangles without wrap-around without area 0. Also there are
(n+1)(m+1)(n+m-2)/2 for those with area 0 but which are not 0 by 0, and (n+1)(m+1) rectangles which are 0 by 0, giving a total of ((n+1)(m+1))*((n+1)(m+1)+n+m+1)/4 for rectangles without wrap=around. For any rectangle, wrap-around introduces the new variable of which border it crosses, which could be horizontal, vertical, none, or both, giving ((n+1)(m+1))*((n+1)(m+1)+n+m+1)

[Who]

Yes I am. I messed up removing the ones with no area, and I also didn't pay attention to the rectangles which did not have area when they didn't wrap around but which did when they did.
So, the new way:

Spoiler: Is it this?

First, let's say a rectangle is defined as 2 sets of x-coordinates and 2 sets of y-coordinates, that makes things much easier.
There are m choices for the first x coordinate (Either edge coordinate would be the same) and 2m-1 choices for the second (2m total points given the wrap-around, one of those is the original point). Of those 2m-1, m-1 can result in a duplicate (the other m choices are wrapped around and do not exist when choosing the first point) and need to be divided by 2. Ultimately, we get m(3m-1)/2 Then we do the same for n, and get (3m-1)(3n-1)mn/4

[Who]

x=6, y=2, z=9

A step-by-step guide:
Step 1: Assume they're all integers.
Step 2: Find the equation which would be the most restriction by said assumption (Eq2)
Step 3: Find the variable which matters the most to said assumption in said equation (y)
Step 4: Make a guess for said variable, start with the lowest possible (Cannot be 1 since they are all greater than 0 and this would force x to be 0, thus start with 2)
Step 5: Find the lowest thing z could be while remaining an integer (Cannot be 1 since that would force x to be 0, thus 4)
Step 6: Find x
Step 7: Check 2,2,4 with EQ1
Step 8: Raise z to the next lowest thing and use it as a new guess (9)
Step 9: Find x
Step 10: Check 6,2,9 with EQ1
Step 11: Check 6,2,9 with EQ3
Step 12: Shoot yourself for doing math this way.

[Who]

x=6, y=2, z=9

A step-by-step guide:
Step 1: Assume they're all integers.
Step 2: Find the equation which would be the most restriction by said assumption (Eq2)
Step 3: Find the variable which matters the most to said assumption in said equation (y)
Step 4: Make a guess for said variable, start with the lowest possible (Cannot be 1 since they are all greater than 0 and this would force x to be 0, thus start with 2)
Step 5: Find the lowest thing z could be while remaining an integer (Cannot be 1 since that would force x to be 0, thus 4)
Step 6: Find x
Step 7: Check 2,2,4 with EQ1
Step 8: Raise z to the next lowest thing and use it as a new guess (9)
Step 9: Find x
Step 10: Check 6,2,9 with EQ1
Step 11: Check 6,2,9 with EQ3
Step 12: Shoot yourself for doing math this way.

Yeah. I think making everything positive made things much easier.

Sort of. Not giving the inequalities would allow you to abuse zero.

[Who]

I tried testing it with smaller numbers and found the answer but could not prove it. All I could get was an unprovable statement that for any subset n which had to add up to something divisible by n k would equal 2n-1, that is, the largest set which could exist of which no subset of length n would have a sum divisible by n would be of length 2(n-1). I then ran into a brick wall when proving that n-1 0s and n-1 1s was the largest set which could exist without a subset of length n the sum of which was divisible by n.

[Who]

Oops?

I have no idea what any of those words mean, and the definitions of each include many more words the definitions of which I do not know, so I am not attempting it.

[Who]

Give the reduced row-echelon form of the following augmented matrix:

Code: Select all

┌             ╷     ┐
│  4 -2  1  3 ╎  45 │
│             ╎     │
│  3 -2 -1 -5 ╎  12 │
│             ╎     │
│  1  0  7 -2 ╎  13 │
│             ╎     │
│ -6  2  8  4 ╎ -16 │
└             ╵     ┘

Spoiler: Equivalent problem for people who are unfamiliar with matrices

You are essentially being asked to solve this system of equations:
4x-2y+z+3w=45
3x-2y-z-5w=12
x+7z-2w=13
-6x+2y+8z+4w=-16

By "reduced row-echelon form", I'm looking for the identity matrix on the left side of the line, so I want your answer in this format:

Code: Select all

┌         ╷   ┐
│ 1 0 0 0 ╎ x │
│         ╎   │
│ 0 1 0 0 ╎ y │
│         ╎   │
│ 0 0 1 0 ╎ z │
│         ╎   │
│ 0 0 0 1 ╎ w │
└         ╵   ┘

Replace

x

,

y

,

z

, and

w

with what you actually got for those variables.

I'm lazy. Can we use calculators?

[Who]

Code: Select all

┌         ╷   ┐
│ 1 0 0 0 ╎ 5 │
│         ╎   │
│ 0 1 0 0 ╎-7 │
│         ╎   │
│ 0 0 1 0 ╎ 2 │
│         ╎   │
│ 0 0 0 1 ╎ 3 │
└         ╵   ┘

[Who]

Either find an integer value for x such that x+3 and x^2+3 are both perfect cubes or prove that none exist.

[Who]

Both of those work.

Spoiler: My proof, which was far more complex than it should have been due to me not thinking of the product being a perfect cube when I first did it

My solution:
For positive x+3s:
If x+3 is a perfect cube then x+3=m where m is a perfect cube, m=n^3 where n is an integer)
x+3=m, x=(m-3), x^2=(m-3)^2=m^2-6m+9, x^2+3=m^2-6m+12=
m^2 is a perfect cube.
The difference between x^2+3 and m^2 is m^2-6m+12-m^2=-6m+12=-6n^3+12
If x+3 is positive then (x+3)^2>x^2+3
The closest cube to n^6 which is greater than n^6 is (n^2+1)^3, thus if x^2+3 is in between n^6 and (n^2+1)^3 then it is not a perfect cube.
The difference between (n^2+1)^3 and n^6 is 3n^4+3n^2+1
Thus if 3n^4+3n^2+1>12-6n^3 then x+3 and x^2+3 are not perfect cubes where n is positive.
Thus if 6n^3+3n^2+3n>11 then x+3 and x^2+3 are not perfect cubes where n is positive.
The minimum n can be is 1 if it is positive, 6+3+3=12>11, thus x+3 and x^2+3 cannot both be perfect cubes where n is positive.
For negative x+3s:
If x+3 is a perfect cube then x+3=m where m is a perfect cube, m=n^3 where n is an integer)
x+3=m, x=(m-3), x^2=(m-3)^2=m^2-6m+9, x^2+3=m^2-6m+12=
m^2 is a perfect cube.
The difference between x^2+3 and m^2 is m^2-6m+12-m^2=-6m+12=-6n^3+12
If x+3 is negative then (x+3)^2<x^2+3
The closest cube to n^6 which is less than n^6 is (n^2-1)^3, thus if x^2+3 is in between n^6 and (n^2-1)^3 then it is not a perfect cube.
The difference between n^6 and (n^2-1)^3 is n^6-n^6+3n^4-3n^2+1=3n^4-3n^2+1
Thus if 3n^4-3n^2+1>12-6n^3 then the hypothesis is false for negative ns
Thus if 3n^4+6n^3-3n^2>11 then the hypothesis is false for negative ns
3n^4+6n^3-3n^2-11=f(x)
f'(x)= 12n^3+18n^2-6n
The critical numbers of that are: n=0, n=.28, n=-1.78
When n=-3 then f'=-144 and f=43, given how it is decreasing, positive, and with no critical numbers of the first derivative, n cannot be less than 3.
Brute force time!
If n is -1 then x is -4 and x^2+3=19=not a cube
If n is -2 then x is -11 then x^2+3=124=not a cube
If n is -3 then x is -30 then x^2+3=903=not a cube
Thus x+3 and x^2+3 cannot be a perfect cube.

Also another way to do it is to prove that x^2+3 can never be a perfect cube for any integer x. I don't understand how to prove that, but it can be done.

[Who]

0

[Who]

Last I checked sin0=0, cos0=1, tan0=0, cot0=undef, sec0=1, csc0=undef.

[Who]

2) Show that if r is positive and r^r is rational, r is either an integer or irrational.

Spoiler:

Defining variables:
x is prime
n is a positive integer
L is a positive integer
m is a positive rational number
c is rational
k is any number
K is any integer
For r=1/x:
x^m will always be irrational, because if x has no factors (other than 1 and itself) there is no number you can multiply by itself n times to reach x. Not sure if that's technically rigorous enough but whatever, it makes sense.
Thus x^1/n will always be irrational
Thus x^1/x will always be irrational
The inverse of an irrational number is irrational.
Thus if r=1/x and x is a prime integer, r^r will be irrational.

For r=1/x^n:
1/x^n=(1/x)^n
(1/x)^n^(1/x^n)=(1/x)^(n/x^n)
n/x^n is rational and less than 1
Thus (1/x)^(n/x^n) is irrational
Thus if r=1/x^n, r^r is irrational

For r=1/ab, where a and b are both different primes raised to an integer power
As proven earlier, a^1/a and b^1/b are irrational
If a^K/L is irrational, for k*a^K/L to be rational, k must be c*a^n/L
a and b are coprime, thus there is no way to express a^1/ab as c*b^n/ab
Thus ab^1/ab is irrational, thus 1/ab^1/ab is irrational, thus if r=1/ab then r^1/r is irrational
The above proof works for more than just a and b, if I knew the notation for a mysterious number of variables then I would have used it, but I don't, but the proof still works.
Thus, r^1/r is irrational when r is the inverse of an integer.

For r=p/q where p and q are coprime and q≠1:
(p/q)^(p/q)=(p^(p/q))/(q^(p/q))
As proven earlier, q^(1/q) is irrational
Since q and p are coprime, q^(1/q)^p is irrational, thus q^(p/q) is irrational,
thus since r^r=p^(p/q)*q^(-p/q), p^p/q must be c*q^n/q to make r^r rational, but p and q are coprime so it can't be.

Thus no r^r where r=p/q where p and q are coprime and q≠1 is rational
Thus no r^r where r is not an integer and rational (not irrational) is rational
Thus if r is rational, r is either an integer or irrational

[Who]

The solution for 1) is exactly the one I have.

I think you're basically right about 2, though there are nicer/more rigorous ways to go about the first parts.

One thing I'm unsure about (not with your proof, but in general):

n/x^n is rational

and less than 1

Thus (1/x)^(n/x^n) is irrational

This is of course true, but if we're going to use that sort of reasoning, I think it's sufficient to observe that

1 < q^(1/q) < 2,

so from mitsi's proof above, q^(1/q) must be irrational for any integer q>1.

I'm not sure if you can prove this without relying on some sort of inequality. My own method from scratch relies on the same:

Spoiler:

Let r=a/b and r^r = c/d and, with a, b, c, and d positive integers and gcd(a,b) = gcd(c,d) = 1. Then r^r=(a/b)^(a/b)=c/d, so a^a * d^b = b^a * c^b. Now, gcd(a^a, b^a) = gcd(c^b, d^b) = 1, so d^b = b^a.

For some prime factor of b, p, where p|b, define positive integers m and n such that p^m || d and p^n || b (meaning that p^m is the highest power of p that divides d). From the definitions of m and n, p^(m*b) = p^(n*a) and b = n*a/m, but we know p^n | b and gcd(a,b) = 1, so p^n | n. Since p^n > n for all p>1, b can have no prime factors, so we conclude that b=1.

That's the unrigorous part?

I was expecting the part about the product only being rational if they were both the same base to be the problem.

n/x^n<1 is easy to prove with basic calculus, your 1<q^(1/q)<2 is far more elegant though. Also, your proof uses almost exactly the same thing with "p^n > n"

[Who]

Let's throw in a new one:

Prove that, for any real constant c, the derivative of ce^x is itself.

For what definition of e^x?

Using the sigma thing:

dcf(x)/dx=cf'(x), thus if the derivative of e^x is e^x, the derivative of ce^c is ce^x.

de^x/dx=dsum(k=0 to infinity, x^k/k!)/dx=sum(k=0 to infinity, dx^k/k!/dx)=sum(k=0 to infinity, kx^(k-1)/k!) = 0+sum(k=1 to infinity(x^(k-1)/(k-1)!)=sum(k=0 to infinity(x^k/k!)=e^x

[Who]

Other definitions:
Inverse of integral(1 to x,1/t):
df^-1(x)/dx=1/f'(f^-1(x))
If f'(x)=1/x then the above equals f^-1(x)

Limit n->infin (1+x/n)^n:
d(1+x/n)^n/dx=n/n(1+x/n)^n-1=1*(1+x/n)^(n-1)=(1+x/n)^n/(1+x/n)=lim((1+x/n)^n)/lim(1+x/n)=e^x/1

[Who]

I wasn't looking for the "by definition" answer for ce^x. I wanted work shown.

Edit:

To clarify, the definition of e^x is not taught in precalculus that way; an approximate value of 2.818 for e is taught before derivatives are taught. I was looking for an answer based on applying the rules of derivatives without that assumption.

What definition do you want us to use then?

[Who]

It should be sec(x)tan(x)+sec^3(x).

He probably made the mistake of forgetting the product rule and thinking d(f(x)*g(x))/dx=f'(x)g'(x) when taking the derivative of secxtanx (giving secxtanx*sec^2(x), when it should be sec(x)tan(x)+sec^3(x))

[Who]

Typo, should have been secxtan^2(x)+sec^3(x)

[Who]

Does "after" mean "directly after" or just "after"?

[Who]

I know it's not my turn, but this came up in discussion at school, and I thought people here might enjoy it.

Two people are drawing marbles from an urn, which contains 50 black marbles and 50 white marbles. The first person draws a marble, and if it's black, he wins. Otherwise, he sets it aside, and the second player draws. Again, if the marble he draws is black, he wins. Otherwise, he puts it aside. The two players continue drawing marbles in this way until somebody draws a black marble and wins the game. What are the odds that the first person to draw will win?

Do we get a calculator? Do we get a fancy calculator which can do for loops?

[Who]

f(x)	f'(x)	int f(x)
sin(x)	cos(x)	?
cos(x)	-sin(x)	?
tan(x)	?	?
cot(x)	?	?
sec(x)	?	?
csc(x)	?	?

Fill in the remainder of the table and show your work.

Spoiler: Hint for the int f(x) column

int_1^a (1/x) dx = ln(a)

f(x)	f'(x)	int f(x)
sin(x)	cos(x)	-cos(x)+c
cos(x)	-sin(x)	sin(x)+c
tan(x)	sec^2(x)	ln|sec(x)|+c
cot(x)	-csc^2(x)	ln|cscx|+c
sec(x)	secxtanx	ln|secx+tanx|+c
csc(x)	-cscxcotx	-ln|cscx+cotx|+c

I don't know if it's standard notation but I use (f(x))' to mean f'(x). Deal with it. Also I put the ' before the parenthesis on a trig function sorta like people do with exponents. Also deal with that.

sin'(x)=lim h-> 0 of (sin(x+h)-sinx)/h = (sinxcosh + sinhcosx - sinx)/h =(sinxcosh - sinx) / h + sinhcosx/h = (sinx(cosh - 1)(/h + cosx = cosx. Do I have to prove the (cosh - 1)/h=0 and sinh/h = 1?
cos'(x) = lim h-> 0 of (cos(x+h)-cosx)/h = (cosxcosh - sinxsinh - cosx)/h = cosx*(cosh-1)/h - sinxsinh/h = -sinx. Same question.
∫sin(x) = antiderivative(sin(x)) = ??? = Ooh shiny the derivative of the negative cosine is the sine, good thing that's a nice coincidence because if I had to actually figure it out without going "ooh shiny nice coincidence" I would have no clue where to start.
Same for ∫cosx
tan'(x)=(sinx/cosx)'=(cos^2(x) + sin^2(x)) / (cos^2(x) = 1/cos^2(x)=sec^2(x)
∫tan(x):
u=cos(x)
-du=sin(x)dx
-∫1/u du = -ln|u|+c=ln|1/u|+c=ln|sec(x)|+c=ln|c*sec(x)| but nobody ever writes it that way because they're lame.

cot'(x)=(cosx/sinx)'=(-sin^2(x) - cos^2(x)) / (sin^2(x) = -1/sin^2(x)=-csc^2(x)
∫cot(x) = ∫(cosx/sinx)dx
u=sinx
du=cosxdx
∫(1/udu)=ln|u|+c=ln|sinx|+c

sec'(x)=(1/cos(x))' = --sinx/cos^2(x)=sinx/cosx * 1/cosx=secxtanx
csc'(x)=(1/sin(x))' = -cosx/sin^2(x) = -cscxcotx

∫sec(x)dx =
...
∫1/cos(x)dx=
u=???
du=???
1/cosx=du/u
f'(x)/f(x)=1/cosx
dy/dx=y/cosx
dy/y=dx/cosx

Fuck it, time for more convenient coincidences. It just so happens that d/dx(secx+tanx) = secx(secx+tanx).

du/u=1/cosx
f'(x)=f(x)sec(x)
u=secx+tanx
du=secx(secx+tanx)
∫du/u=ln|u|+c=ln|secx+tanx|+c

∫cscxdx =∫1/sinx dx
If ∫secx couldn't be done without convenient coincidences, ∫cscx probably can't be either.
u=cscx+cotx
-du=cscx(cscx+cotx)dx
-∫du/u=-ln|u|+c=-ln|cscx+cotx|+c

[Who]

You sure it's tan(x) which is off and not cot(x)?
I did cot(x) right in the work, but then put it wrong in the table. It should be:

f(x)	f'(x)	int f(x)
sin(x)	cos(x)	-cos(x)+c
cos(x)	-sin(x)	sin(x)+c
tan(x)	sec^2(x)	ln|sec(x)|+c
cot(x)	-csc^2(x)	ln|sin(x)|+c
sec(x)	secxtanx	ln|secx+tanx|+c
csc(x)	-cscxcotx	-ln|cscx+cotx|+c

[Who]

There are infinitely many prime numbers and infinitely many even numbers, though, and their definitions aren't mutually exclusive, either.

They are for n≠2.

[Who]

No new puzzles... Anyway, here's one (it really is a puzzle):

There is a circular highway, a car and finitely many gas stations. It takes 1 unit of fuel to travel around the highway, the sum of the amount of fuel stored at the gas stations is 1 unit. The car is empty. You may choose where the car is at the beginning. Is it possible to travel the hignway around?

Yes*.
...
What?
You didn't say anything about having to prove it.

*Definitely (Can think of a justification I accept but too lazy to formally prove) for rational amounts of fuel/station, almost definitely (It shouldn't make a difference) for irrational amounts of fuel/station.

General argument:
Assume all of the n stations have the same amount of fuel.
If all the stations have different rational amounts of fuel then you can just replace one station with more stations all at the same place, thus the having the same amount of fuel doesn't matter.
The maximum minimum distance between two stations is 1/n, the distance you can go on one station's worth of fuel is 1/n. That is, the way to place stations such that you maximize the distance between the two closest stations is to place them at even intervals. If you were to move any two stations further apart than 1/n units from that setup you would be moving two more closer together than 1/n units from that setup.
The exact same logic applies to the maximum minimum distance between three stations
The exact same logic can be repeated n times.

[Who]

Pascal's triangle/the binomial formula.
Sum(k=0 to n, n!/(k!(n-k)!) *x^k *y^(n-k))

[Who]

New puzzle. I don't know the answer despite trying to solve it, I think it requires more advanced math than I currently know.

A velociraptor is 40 meters away from you when you see each other. It accelerates towards you at 4 m/s^2 (The magnitude of this acceleration remains constant and the direction is always pointed towards you). You, thinking "Fuck reality, I'm pretty sure that I can live according to the rules of this math problem provided I don't run away", start running at 6 m/s in a straight line perpendicular to the line from your starting position to the velociraptor's starting position.
Assume that neither you nor the velociraptor gets tired.

Does the velociraptor catch, kill, and eat you? If it does, how long does that take? If it doesn't, what is the closest it ever gets to you?

[Who]

Sorry if I worded it ambiguously: Mith's interpretation about acceleration is correct. It is the acceleration, not the velocity, which is always pointed towards you.

The question is though: How do you prove that it will spiral around and never reach you, and how close does it get at the closest point?

[Who]

Just by looking at the problem, I suspect that it might be 512.

There is definitely a solution where 512 is the mass of the largest stone, so the actual answer can be no more than this.

It isn't 512.
The logic which leads to 512 leads to 8 as the minimum largest mass of a similar 4-stone set, but 1,3,5,7 works.

[Who]

1 + 7 = 3 + 5 = 8

...
, I'm incapable of doing basic arithmetic.

Probably 512 then.

[Who]

The following is not always correct, sometimes certain things are positive when they should be negative. I'm assuming that this is what you meant though, because I'm pretty sure that otherwise it's blatantly impossible unless you make the function piecewise-defined.

In terms of:	sin(x)	cos(x)	tan(x)	cot(x)	sec(x)	csc(x)
sin(x)=	sin(x)	√(1-cos2(x))	√(tan2(x)/(tan2(x)+1))	√(1/(cot2(x)+1))	√(1-1/sec2(x))	1/csc(x)
cos(x)=	√(1-sin2(x)	cos(x)	√(1/(tan2(x)+1))	√(cot2(x)/(1+cot2(x)))	1/sec(x)	√(1-1/csc2(x))
tan(x)=	√(sin2(x)/(1-sin2(x)))	√((1-cos2(x))/cos2(x))	tan(x)	1/cot(x)	√(sec2(x)-1)	1/√(csc2(x)-1)
cot(x)=	√((1-sin2(x))/sin2(x))	1/√((1-cos2(x))	1/tan(x)	cot(x)	1/√(sec2(x)-1)	√(csc2(x)-1)
sec(x)=	1/√(1-sin2(x)	1/cos(x)	√(tan2(x)+1)	1/√(cot2(x)/(1+cot2(x)))	sec(x)	1/√(1-1/csc2(x))
csc(x)=	1/sin(x)	1/√(1-cos2(x))	√((tan2(x)+1)/tan2(x))	√((1+cot2(x))/cot2(x))	1/√(1-1/sec2(x))	csc(x)

Most of these could be simplified but you didn't say they had to be so I won't.

Edit: Typo, sec(x) is 1/√(1-sin²(x) not 2/√(1-sin²(x).

[Who]

Let f(x) be whatever it said in the table
For all the ones which weren't already filled out:
sin(x) = f(x) when there is an integer k for which 0<x+2πk< | -f(x) for all other values of x.
cos(x) = f(x) when there is an integer k for which -π/2<x+2πk<π/2 | -f(x) for all other values of x.
tan(x) = f(x) when there is an integer k for which 0<x+πk<π/2 | -f(x) for all other values of x.
csc(x) = f(x) when there is an integer k for which 0<x+2πk< | -f(x) for all other values of x.
sec(x) = f(x) when there is an integer k for which -π/2<x+2πk<π/2 | -f(x) for all other values of x.
cot(x) = f(x) when there is an integer k for which 0<x+πk<π/2 | -f(x) for all other values of x.

And the 2nd 3 rows were almost entirely just copy-pasted from the first 3. Sometimes I'd delete a 1/ rather than adding a 1/.

[Who]

Using four 4s and any unary or binary operations, can you make expressions that equal each integer between 1 and 100? Or at least as many as possible?

Example: 1 = 4/4 + 4 - 4

Define "any".
Any that a layman would know? Any anywhere which already exist? Are we free to define our own?

If it's the first, I don't know give me a list of functions. If it's the 2nd, yes. If it's the 3rd, yes and it's really easy.

[Who]

Defining your own is not allowed, because then you could just say 4$4 = 67, or something. I guess I mean ones which are standardised, so probably the second.

Well actually I was planning to define f₂(x)=2/4, f₃(x)=3/4, etc.

Anyway:
I shall use the unary operators: sin, cos, tan, inverse (As in, 1/x), floor, ceiling, ! (factorial), gamma, - (negate)
And the binary operators: +, -, /, *

2 = 4/4+4/4
3 = 4-4+floor(-4sin(4))
4 = 4-4+ceiling(-4sin(4))
5 = 4-4+ceiling(4tan(4))
6 = 4 - 4 + 4!/4
7 = 4+4 - 4/4
8 = 4 + 4 + 4 - 4
9 = 4+4+4/4
10 = 4 - 4 + 4 + gamma(4)
11 = 4 + 4 + floor(4sin(4))
12 = 4 + 4 + ceiling(4sin(4))
13 = 4 + 4 + ceiling(4tan(4))
14 = 4 + 4 + 4!/4
15 = 4 * 4 - 4 / 4
16 = 4 * 4 + 4 - 4
17 = 4 * 4 + 4/4
18 = 4 * 4 + floor(-4cos(4))
19 = 4! - 4 - 4/4
20 = 4! - 4 - 4 + 4

And I'm too lazy to go any further for now. Do you have a problem with how I did any of these?

[Who]

An easy calculus problem:

You type in a positive number in a calculator and hit the square root key over and over. Each time you press the square root key, the answer produces becomes closer to what number? (That is, for any positive x, what is the limit of x^(1/(2^n)) as n approaches infinity?)

Speaking as someone who has done this when I was much younger, 1. If you want a more rigorous proof then I can supply one.

[Who]

proof
take x ^ ((.5)^n) as n-> inf then we have x^0 = 1

Counter proof (Uses same logic to reach wrong conclusion):
Take floor(sin(x)/x) as x -> 0 then we have floor(1)=1.

But actually floor(sin(x)/x) as x -> 0 = 0.

Thus your proof is missing something.

[Who]

That is, for any positive x, what is the limit of x^(1/(2^n)) as n approaches infinity?

As n→∞, 2^n→∞.
So (1/(2^n))→0.
Therefore x^(1/(2^n))→1. (x > 0)

Is it not as simple as that?

lim(f(g(x))) doesn't always equal f(lim(g(x))). Refer to my counterexample.

[Who]

The limit is as n tends to infinity. It's not a limit in x, so... I don't see the problem.

It doesn't matter what letter you use. lim(f(g(n))) doesn't necessarily equal f(lim(g(n)). In this case f(n)=x^n and g(n)=1/(2^n).
The limit as n approaches 0 of sin(n)/n=1. Floor(1)=1. But the limit as n approaches 0 of floor(sin(n)/n) = 0, not 1.

[Who]

1)

b1 d2 f3 h4 c5 a6 e8 g8

2)

Impossible. The 8 queens would all have to be in separate rows/columns, there are 8 columns, thus there is one queen in each row and one in each column. If there are no other pieces on a board then a queen controls every square in her row and column, thus the queens must control the entire board.

Also, in case anyone wants, a solution to the original problem:

b1 d2 f3 h4 c5 a6 e8 g7

Yes I did just solve the original problem then alter the solution so that two were attacking each other.

[Who]

@Who: Very nice.

There are, of course, many more possibilities for 1. As for 2, can it be done with 7 queens?

For 1, yes there are. I think there are also solutions which aren't just solve the original then change it, I remember that when I originally solved the original (Which was quite a while ago) I kept running into problems with the last one.
For 2, yes. Any solution to the original, minus one queen. For example,

b1 d2 f3 h4 c5 a6 e8

.
Because they'll all be in separate rows/columns but there will be one square which isn't in a controlled row or a controlled column (The slot where you would place the final queen). Also, it's only solutions to the original minus a queen because if you put a queen in that square then it's a solution to the original.

Too lazy to do knights and bishops because I don't have a chessboard to hand to allow me to visualize it and I was doing the queen problems on this flash game. (Which is just the original problem. I played it a while ago and beat it, and the solution was easy to remember when solving these)

[Who]

New puzzle.
I'm sure you're all familiar with the labyrinth guard puzzle. Unfortunately, after prisoners kept escaping the management was forced to hire a new guard. The result was the following situation:

Spoiler:

The new guard did so well that they fired the other two. There is now just one guard, and that guard can either lie or tell the truth whenever he wants, he isn't bound to one specific kind of truthfulness. And he'll stab you if you ask a tricky question. "But what is a tricky question?" you might ask. If you asked him that he'd stab you, because any question which isn't a yes-or-no question counts as a tricky question. Also, any question where the guard is incapable of giving an answer which is true or a lie (I'm too lazy to think of one, but basically no questions where the guard's only option is to answer you with a paradox) counts as a tricky question.

You have a question for the guard, the question is: "Will going through this door kill me?". You can ask the guard as many questions as you want, and for any given question the guard can tell the truth or lie. What question(s) do you need to ask to find out?
Logical operators (And, Or, Xor, etc.) are allowed and encouraged. Self-referential questions are allowed provided they don't break any of the rules on tricky questions.

[Who]

Oh, that might have to be "You're

evaluating these two statements truthfully/falsely and

answering this question truthfully/falsely".

EBWO(¬DP): Just checked that - assuming that the guard evaluates each OR statement, inverting both results if he's being untruthful, then evaluates the AND (with those modified results) and inverts that result if he's being untruthful, then the solution works. I don't think that's the intended reading of the question; I might have to keep thinking on it.

If he's lying he doesn't invert anything until right before he answers the question. Otherwise it's possible without any self-reference simply by asking "(Will the door kill me OR Will the door kill me) AND (Will the door kill me OR Will the door kill me)?" (Invert the or results, invert the and results, they cancel out)

Sorry if it was unclear: The guard doesn't lie about individual phrases in the question, they just evaluate the entire question and then, if they're lying, they flip the entire result. If you ask them a hypothetical question about what they would answer, that's where they can lie in an individual part (Because they could hypothetically lie).

[Who]

You need something like, "Is it true that either going through this door will kill me, or your answer to this question is a lie, but not both?"

That works. (With logical operators you could have just used exclusive or, but your method of asking does the same thing)

[Who]

If a tree falls in the forest, what color is it?

From what height was it falling?

Assuming it was falling from a random height on (0,the farthest away at which it will still fall to earth), it's probably black.

[Who]

Bring 1000 donuts 1000/6 meters, dump 2000/3, go back. Repeat twice (Except ditch the last "go back"). He now has 2000 donuts @ 1000/6 meters. Bring 1000 donuts 1000/3 more meters, dump 1000/3, go back. Bring another 1000 donuts 1000/3 meters, he now has 1000 donuts@500 meters. Deliver 500. Stay.

Best I can think of without arm length.

I can do better.

Bring 1000 donuts 200 meters, dump 600, go back. Repeat. Bring the final 1000 to 200 meters. He now has 2000 donuts@200 meters. Now bring 1000 donuts 1000/3 more, dump 1000/3, go back, bring the final 1000, he now has 1000 donuts@1000/3+200 meters, deliver, he delivers 500+100/3 donuts.

That's the best I can think of.

[Who]

Actually I think that that's the best possible.

When starting with 2000+ donuts it costs 5 d/m to deliver.
When starting with 1000-1999 it costs 3 d/m to deliver.
When starting with 1000 or less it costs 1 d/m.
This is true regardless of how far he goes per trip. For example, if he goes 1m/trip then on the first trip he would bring 1000 1m, dump 998, go back, bring the next 1000 1m, dump 998, go back, bring the next 1000 1m, and have 995. (Or, generalize it to x m/trip, he brings 1000 xm, dumps 1000-2x, goes back, repeats, brings the final 1000 but eats x to bring them there, for a total of 5x eaten).

Thus, provided he wastes no donuts on trips so long that he could divide the trip into 2, one of which he starts with less than the threshold for the next lower cost, he will be able to bring 1000 donuts 1000/5+1000/3 m, and then go the rest of the distance on those 1000, delivering a total of 1000-(1000/5+1000/3).

If the deliveryman needs to get back:
On the first trip to any point, drop a donut every meter to eat on the way back, it now costs:
6 d/m when starting with 2000+
4 d/m when starting with 1000-1999
2 d/m when starting with <1000

Go 1000/6m then 1000/4 meters then he's gone less than 500m and cannot get to your house. I don't think that the deliveryman can get home, unless someone comes up with a more efficient method.

[Who]

@Dybeck
Just to confirm, you wouldn't know the number if the friend didn't answer the last question?
It seems as if that is the implication, but my math textbook recently drilled into my head that if≠iff.

Assuming you meant "iff":
Given how you asked question 4, he must have told you the number was 50 or less, making it greater than 50.
Given how you could have known, he must have answered yes to question 3.
So, based on what he told you:
Number<=50
Number is a perfect square
Possibilities: 9,16,25,36,49
If he answers no to q2: 9,25,49
If he answers yes to q2: 16,36
q4 would only help you if the answer to q2 was "yes", meaning he answered "yes" to q2.

Meaning the number is:
A perfect square. (9,16,25,36,49,64,81,100)
Larger than 50. (64,81,100)
Not a multiple of 4. (81)

Leaving: 81.
Your friend made a good choice. That was my favorite number back in elementary school.

[Who]

There are infinite ways in which logical statements can be combined. (Proof: For any logical statement S, one can add another logical connector and another variable)
For any combination of logical statements A, there is a tautology A or ~A.
For any combination of logical statements A, there is a contradiction A & ~A.
Thus, there are infinite tautologies and infinite contradictions.

[Who]

New puzzle:
There are 3 princesses, all sisters. The eldest always lies, the youngest always tells the truth, and the middle answers randomly. You are going to marry one. You do not want to marry the middle sister, because with the other two you will always know where you are. You have to decide which one to marry by asking one of them one yes-or-no question. You may not ask an unanswerable question.

What question do you ask to find a sister who is not the middle sister? (Note: The middle sister just answers "yes" or "no" randomly. No forcing her to give you info.)

Fun fact about this puzzle: When I first saw it I could not figure it out and looked at the answer. And then I facepalmed because it was the exact same as another puzzle I had already solved but was not thinking about.

[Who]

Can you tell which princess is which before you ask the question, or is your question posed to a random princess?

There are 3 princesses: A, B, and C.

You can ask any one you want but you don't know which is which.

The goal is to identify one of them who isn't the middle sister.

[Who]

That works.

Another way of doing it is to treat it just like the 3 gods puzzle. ("Would you say 'yes' if I were to ask you if B was the middle one")

Despite having already solved the 3 gods puzzle when I first encountered it I deemed the 3 princesses puzzle impossible and looked at the answer. That was funny.

[Who]

You just have to marry a non-middle, right? You don't have to actually figure out which one the liar is? Ask them anything you know the answer to. Are you female? Is 3 greater than 4? Then just pick the princess that gave a unique answer. You won't know which of the other two is the random one, but you'll successfully avoid her and also know if your bride is the truther or the liar.

But you don't get one question to ask all 3, you get one question to ask one.

[Who]

So they do not know each other's guesses?

Also, do they know how many colors there are in advance? And do they know that there has to be at least one of each color? Does there have to be at least one of each color?

[Who]

Here's another prison one:

Suppose a prison has n prisoners on death row. Once per day, a random prisoner is picked (and the same prisoner can be picked more than once) and sent alone into a room, which is not visible from any of the prisoners cell, where there is a single light bulb, which randomly starts on or off. The prisoner may turn the light on or off or do nothing, as they wish. The prisoners may confer in advance, but may not communicate thereafter except through the lightswitch. At any time, a prisoner may assert that every prisoner has visited the room at least once. If he is right, all the prisoners are free; if he is wrong, all the prisoners are executed. What is the fastest way for the prisoners to win their freedom through this game?

There's probably a faster way, but here's one way:

Spoiler:

Let us designate one prisoner as prisoner zero. He will be the one to declare that every prisoner has visited the room at least once, and he will be the only one to switch the lightbulb off.
"a prisoner" means any prisoner not prisoner zero.
If a prisoner walks in and the light is off and he has not switched the light on twice, he switches it on.
If a prisoner walks in and the light is on or he has switched the light on twice, he leaves it alone.
If prisoner zero walks in and the light is off, he leaves it alone.
If prisoner zero walks in and the light is on, he switches it off.

When prisoner zero has walked in and seen the light switched on 2n-2 times, he knows everyone has been in there at least once.

(Everyone except one person goes in there twice because otherwise the random starting point screws with things)

[Who]

Why not just have the first prisoner turn the light on, if it's off, and leave it alone if it's not, but count it as his "switching on" attempt? Then it doesn't matter what state the light starts as. So, each prisoner can turn the light on once, and leave it alone after that. Then, prisoner zero can count n-1 times, instead.

And if the first one to go is prisoner zero, he can just turn it off, without counting this as one of the n-1 times.

Good idea. That would work.

I think the first version of this I saw was one where it wasn't once a day, it was once every random amount of time, so nobody ever knew if they were the first prisoner.

[Who]

Because the light hasn't been turned on yet. The first.person turns off the light if it's on. If the light is off after n turns, everyone has visited.

And what happens if:
Person A visits room.
Person B visits room twice, is counter, switches it on.
Person C visits room, switches it off to mark that they visited.
Person A visits room again, sees it off, assumes that they are counter, ERROR ERROR EVERYONE DIES!

[Who]

What does ' mean?

[Who]

v₂=a/((t-t₁e^(d/a))

But then integrating that is a nightmare.

[Who]

You end up with the differential equation y'=a/(x-t₁e^(y/a)). How the fuck do you solve that?

[Who]

Wolfram alpha says it's y(x) = (a (t W(-(e^(-c_1/t) x)/t)+c_1))/t. Where W(x) is the product log function, the inverse of y=xe^x

[Who]

Maybe I'm being an idiot, but it looks to me like both snowploughs are moving at the same speed in the same direction (a is a constant, k is the same for both of them when all of the snow is cleared), so they won't collide. It's stated that both snowploughs clear all of the snow (each), and I can only imagine that they can't clear all the snow if they crash - and if all of the snow is cleared, then they both move at the same speed at that point, so neither one can collide with the other.
... I just KNOW that there's something wrong with what I just said, and someone will point out the extremely obvious gap in my logic and I'll feel like an idiot.

Let's say that a=1, k=1 and after 4 seconds the first snowplough sets off. When it starts, it's moving at .25m/s. But it clears all the snow. And now let's say that after 1 second, the second snowplough sets out. When IT starts, it's moving at 1m/s. And it only speeds up as it gets closer.

Also, quilford's hints aren't helping, the differential equation still looks like a nightmare to solve. Maybe it's because I don't know about differential equations. Yep, I'm going to blame that.

Let's try guesstimating:
If (t2-t1)>t1, they'll never meet each other. Or, if t2>2t1. So, we know that there needs to either be a √(2t1-t2) or a log(2t1-t2).
I don't think a matters, but I could be wrong.
As t1 gets really big, the time it takes them to meet gets really small.
k doesn't actually matter.

I'm gonna go with:
t2*log(2t1-t2)/t1

Random guesses ftw? Or am I wrong? Probably the latter.

[Who]

The derivative=e^(-d/a)(1-vt/a)
...
...
Don't judge me if you saw all those edits.
...
...
...
Ok I must have fucked up because I just got that they meet at a meters.


Alternate puzzle for anyone who wants something easy:
What is the largest number which cannot be expressed as 5k+7l where k and l are both natural numbers?

[Who]

What is the largest number which cannot be expressed as 5k+7l where k and l are both natural numbers?

I can't be bothered with an elegant proof right now, but aren't there infinitely many such numbers by the logic that there are infinitely many primes? Adding two composite numbers together gets you another composite number if I'm not mistaken (and 5(1)+7(1)=12, which is composite). I don't think any prime number can be expressed as 5k+7l with k and l both natural.

17 is prime and can be expressed as 5*2+7.
This theory of yours that composite+composite=composite. It is blatantly wrong. If it were true, more than half of all numbers would be prime. (For any prime p for any natural number c<p, c or p-c must be not-composite)

[Who]

23.

for (k,l) multiples, 24 is (2,2), 25 is (5, 0), 26 is (1, 3), 27 is (4, 1) and 28 is (0, 4). Every other number is one of those plus 5c, where c is a natural number.

You could possibly do a proof about how 7l ends in (0, 7, 4, 1, 8) for l=0..4, and from there you can get all 9 possible ending digits by adding some number of 5s.

That is correct. Personally I did it with 0,7,14,21,28 and only looked at the remainder mod 5 but your way works too.

[Who]

A new puzzle:
Let c(n) be the following recursively defined function:
c(1)=1
For any natural number n:
c(2n)=c(n)
c(2n+1)=(-1)ⁿc(n)

Find sum(n=1 to 2013: c(n)*c(n+2))

[Who]

I can't offer a proof, since I used MatLab, but 1?

Are you sure you didn't make an error somewhere when entering in the problem? Because I have a different result I think I can prove.

It might be 1. Or -1.

I can't remember. Although I solved it on the Putnam last year :p.

I solved it while practicing for the Putnam this year.
You don't remember the proof? It's so elegant and shiny and nice.

Though I suppose if you didn't waste an hour in the wrong direction you may not see it that way.

[Who]

I probably did have an error, since I'm now getting -1. That said, I'm eager to see the proof.

So prove it.

[Who]

One definitely exists.
Are we allowed to use calculators/computer programs, or do we have to find it by hand??

[Who]

New problem, should be possible with only knowledge of basic arithmetic/algebra:

Prove that the sequence (1+1/n)^n is non-decreasing without using calculus.

[Who]

Don't shoot.
Gray tries to shoot the perfect shot.

2/3 of the time you get another shot.

If he misses, you still get another shot.

[Who]

Rewrite as (n+a)^3-(n-a)^3=n^3
Giving 6an^2+2a^3=n^3
Thus n is even.

Too lazy to do a rigorous proof but basically assume a is odd, n^3 is obviously 0 mod 4, any odd number is the same mod 4 when cubed, (n-a) cannot be the same as (n+a) mod 4, thus a cannot be 1 or 3 mod 4, thus a is even.

[Who]

New puzzle:

Find, with proof, the maximum on the interval 0≤y≤1 of ∫₀^y √(x⁴+(y-y²)²)dx
(That's the integral from 0 to y of √(x⁴+(y-y²)²)dx. I'm not entirely sure what the standard notation is for a definite integral when typing)

No using calculators/mathematica/wolfram alpha/matlab.

[Who]

It is not.

[Who]

y-y^2 is constantly increasing over [0,1)

Except it isn't. d(y-y²)/dy=1-2y, which is positive on [0,.5), negative on (.5,1] and has a zero at y=.5.

Thus, the function y-y^2 (And by extension, (y-y^2)^2) has a max at y=.5, but then the integral part makes the whole thing increase with y up until y=1, making you still unsure of the answer.

[Who]

@Fish
I don't know analysis, but I'm gonna have to go with "You look as if you probably know what you're doing".

I think this works? EDIT: Added a step.

Obviously,

x² ≤ √(x⁴ + a²) ≤ x² + a, where a ≥ 0.

Now if we consider a = (y − y²), we can write

x² ≤ √(x⁴ + (y − y²)²) ≤ x² + (y − y²), for 0 ≤ y ≤ 1.

Thus integrating gives

y³/3 ≤ ∫₀^y √(x⁴ + (y − y²)²)dx ≤ y³/3 + y(y − y²), for 0 ≤ y ≤ 1.

Both sides of the inequality are increasing between y = 0 and y = 1, and are equal at y = 1.

So ∫₀^y √(x⁴ + (y − y²)²)dx is maximised (= 1/3) on the interval at y = 1.

Yep.

That was my method, except with the minor difference in that instead of using √(x⁴+a)≥x² I just proved that ∫₀^y √(x⁴ + (y − y²)²)dx=1/3 at y=1. (Which I had already proved before coming up with the more important inequality. I had suspected that it was almost certainly y=1 for quite some time before I discovered the actual proof).

I wonder if Who has anything nicer, heh.

If I can't do a problem I post, I'll generally say so.

[Who]

Since all ps are positive, no p is bigger than 1.
Combine all those logs and we get log(product(n*p_i))
Via the theorem of the means, the nth root of the nth root of (product(np_i)) ≤ sum(p_i)=1. (Each element you are taking the mean of is np_i)
Raise both sides to the n, thus product(np_i) ≤ 1
Thus log(product(np_i))≤log(1)=0
Thus, the original question.

b)So I wasn't supposed to assume that?¹

a) again:
Maximize log_enp₁+... on p₁+p₂+...=n
Using lagrange multipliers:
1/p_i=λ*1
p_i=1/λ
Therefore p₁=p₂=....
Giving np₁=1
Giving p₁=1/n
Evaluate at all ps equal 1/n gives log(1)+log(1)+...=0+0+...=0.
Is it a max or a min?
Evaluate at any other point. For example, p₁=1/2, all other ps = 1/(2n-2)
log(n/2)+(n-1)log(1/(2n-2)=log((n/2)/(2n-2)^n-1)) (n/2) is obviously less than (2n-2)^n-1 for all n>1, thus (n/2)/(2n-2)^n-1)<1, thus log((n/2)/(2n-2)^n-1))<0, thus the solution everything = 1/n was a max, thus the max occurs at 0.

b)
Combine all the logs, thus we get log(product(np_i))≤0
Raise e to both sides, product (np_i)≤1
Bring the n out, n^n*product(p_i)≤1
nth root and divide by n, ⁿ√(product(p))≤1/n
Replace 1 with sum(p): ⁿ√(product(p))≤sum(p)/n

Note how in part a, nothing relied on p being less than 1, and the same logic holds if you replaced all the 1s with sum(p)s and all the 0s with log(sum(p))s.

¹I would like to point out that the far more sexy proof method is to prove TOTM with more basic methods and then to prove a with TOTM.

[Who]

Sorry, Who, but part b) doesn't ask you to prove the inequality with the sequence from a) but rather a different one altogether.

I proved that if the first inequality held then the second inequality held.

[Who]

Note how in part a, nothing relied on p being less than 1, and the same logic holds if you replaced all the 1s with sum(p)s and all the 0s with log(sum(p))s.

You mean here? Maybe you could elaborate a bit, because I don't think your reasoning is easy to follow.

No, I mean everything before it. That just showed it held for sum(p)>1

[Who]

That just showed it held for sum(p)>1

Which is not possible, because sum(p) is defined as being equal to one.

Come on. You've gotta be clearer in your setting out. I bet that if this were a question on a test, you would have received only partial credit.

If I were asked to prove it on a test I'd put more effort into rigor. Also depending on how the question was worded I might just use Cauchy's proof, which is a far more elegant proof of the theorem of the means.

But, since you want me to actually copy-paste the logic and replace "1" with "sum(x)", here:

Let S be the sum of all the xs.
Maximize log_enx₁+... on x₁+x₂+...=S
Using lagrange multipliers:
1/x_i=λ*1
x_i=1/λ
Therefore x₁=x₂=....
Giving nx₁=S
Giving p₁=S/n
Evaluate at all ps equal 1/n gives log(S)+log(S)+...=nlogS or log(s^n)
Is it a max or a min?
At the point (S-(some arbitrarily small number), (some arbitrarily small number / n-1), etc.) the sum of the logs becomes some extremely small negative.
Thus it's a max.
Combine all the logs, thus we get log(product(np_i))≤log(S^n)
Raise e to both sides, product (nx_i)≤S^n
Bring the n out, n^n*product(x_i)≤S^n
nth root and divide by n, ⁿ√(product(x))≤sum(x)/n

Happy now?

[Who]

That doesn't really work though.

Welcome to the world of ideal mathematics; where gamblers have infinite funds, cows are spherical, and every surface is frictionless!

You're thinking of theoretical physics. Theoretical math has stuff FAR more fucked up than that.

[Who]

I don't know if the craps problem was answered correctly, or if my proof of Quilford's was acceptable, but nobody's posted here in a while, so...

New puzzle:
Prove or disprove: For any positive integer n, floor(√n+√(n+1)) = floor(√(4n+2))

[Who]

√(4n+1) < √n+√(n+1) < √(4n+2) iff
4n+1 < 2n+1 + 2√n(n+1) < 4n+2 iff
4n^2 < 4n(n+1) < (2n+1)^2 iff
4n^2 < 4n^2 + 4n < 4n^2 + 4n + 1, which is true

Taking floors of the original, floor(√(4n+1)) <= floor(√n+√(n+1)) <= floor(√(4n+2)) (*)
But 4n + 2 is never square (because even numbers square to multiples of 4, and odd numbers to odd numbers). So floor(√(4n+1)) = floor(√(4n+2)) for all n. So the left and right sides of * are equal, and so is the middle is as well.

This is correct.

---

There are 3^n ways of colouring the first n numbers red, blue and green. Do more of those colourings have an even or odd number of red numbers?

Even.

You didn't say prove.

If a>b then there are more ways of coloring with b red numbers than with a red numbers (Because not-red can be blue or green while red can only be red).
So there are more ways of coloring 0 red than 1 red, more ways of coloring 2 red than 3 red, etc. all the way up to more ways of coloring (3^n)-1 red than 3^n red.
When you add up all the inequalities, you get that the sum of the ways of coloring with an even number of red numbers is greater than the sum of the ways of coloring with an odd number of red numbers.

Alternate proof, which makes assumptions based on the phrasing of the question:
The question implies that it's the same for all n. For n=0, we have 1=red, 1=blue, and 1=green. 1=blue and 1=green both have an even number of red numbers, 1=red has an odd, therefore there are more with an even number for n=0, by the wording of the question it's the same for all, therefore even for all.

[Who]

The provider of easy math problems to the rescue!

A student was asked to solve sqrt(x + 16) for x = 4 and gave the incorrect answer of 6. Find the mistake the student made, disprove the mistake's validity, and correctly solve sqrt(x + 16) for x = 4 in simplest form.

The student incorrectly assumed that √(a+b)=√(a)+√(b), which is incorrect for a,b>0.
Proof:
Create a right triangle of leg lengths √a and √b.
√a+√b>√(a+b) because the sum of the lengths of two of the sides is always greater than the third.

√(4+16)=√(20)=√4*√5=2√5

[Who]

Back to the fun stuff:
If x=(x₁+x₂+...+x_n)/n and 0<x_i<π for all i, prove that:

Product(sin(x_i)/x_i)≤(sin(x)/x)^n

(i covers every value from 1 to n. Not sure how to express that without while typing)

[Who]

sin(x) = ∑_{k=0 to ∞} [ (-1)ⁿx²ⁿ⁺¹/(2n+1)! ]

[Who]

It's time to go back to pure logic:
Everyone has heard of the classic labyrinth puzzle, two doors each guarded by a guard, one door leads to freedom the other leads to death, one guard always tells the truth and another always lies.
But too many prisoners were escaping, so they hired a third guard to manage the other two:

Spoiler:

He will stab you if your question involves any of the following words:
"Would"
"Say"
"If"
"Other"
"You"
(And any conjugation/pluralization of any of the above words)
He will also stab you if you ask:
A self-referential question
A question in which one part references a different part
A question referring to a future/past/hypothetical situation/question/answer/statement.

The stabby guy doesn't speak and will stab you if you ask him any question at all. (You know which one's the stabby guard as he's not guarding either of the doors but rather is just there watching everything)

You get to ask one question to one guard, and then you choose a door. What question do you ask to figure out which door is which?

[Who]

Can I use XOR?

Yes.
(In case anyone is wondering what the thing about referencing different parts was, that was an attempt to ban exceedingly fancy and stealthy self-references. Also there's one thing I'm not sure if it's possible to make a solution out of which gets banned by the other-part-reference thing)

[Who]

Also, I forgot to mention: Only yes-or-no questions. Actually other binaries are accepted (Which door?). But only questions with binary answers.

And does the liar negate statements piecewise, or as a whole? What would his response be to the question: "is 1+1=2 XOR is 1+1=3?"?

If it's possible to force the liar to consider a question in parts, this is solvable. But that trick is just the parallel form of the self-reference question, where the double negation occurs in series.

As a whole.
The answer the liar would give to: "Is 1+1=2 XOR is 1+1=3" would be "no".

If you have a question which refers to the truthfulness of the entire statement, they will evaluate that as if it is a lie and then negate it at the end, so that works, but if you have that you get stabbed, so it doesn't.

Or else, "did you know there's free beer through the good door?" and follow him through.

Please. These are trained professionals. They won't fall for that.

Shoot one of them in the foot with a gun. If he says "You shot me!", he's the truth-teller, and you can ask him which door to go through. If he says "You didn't shoot me!", then he's the liar, and you can ask the other one. If the stabby guy tries anything, shoot him too.

They only speak when spoken to. Also, they are bullet-dodging ninjas and will kill you if you try to shoot them.

[Who]

"Could your partner have answered "yes", should I have asked him wether the first door was the good one?"

You get...
...
...
Stabbed for asking a question about a hypothetical situation. (You were asking about the hypothetical situation where you asked the other guard if the first door was the good one)

Question from an idiot: How does the answer no come from "1+1=2 XOR 1+1=3?" I don't get it - one's true, one's false, therefore it's a yes.

Yes it is, and then he's a liar so he lies and says no. If you asked the truth-teller "1+1=2 XOR 1+1=3", he would say yes. But the question was "what would the liar say".

1+1=2: Y
!+1=3: N
Y XOR N: Y
~Y: N
Thus he says no.

"Does the liar guard the door to freedom?"

Edit:
If the guard answers "No" it's either the honest guard guarding the door to freedom or the lying guard guarding the door to freedom and lying about it, so you choose that door.
If the guard answers "Yes" it's either the honest guard guarding the door to death or the lying guard guarding the door to death, so you pick the other door.

You get...
...
...

to find out that PLOT TWIST! ALL OF THEM LEAD TO DEATH!

freedom. Congratulations on solving the puzzle.

[Who]

I believe what you meant was:

This was a real $16,000 question on Who Wants to Be a Millionaire?:

Which of these numbers is the square of the third number of a pythagorean triple?

1. 16
   
2. 25
   
3. 36
   
4. 49

The contestant used his Ask the Audience lifeline and went with the most popular answer of 16, which is wrong. The incorrect answer can be replicated with a variation on something I've asked to disprove before in this thread. Where can the disproof I asked for then be found? What is the correct answer to the question asked on Millionaire?

The correct answer is B, 25, because (3,4,5) is a pythagorean triple.

(Wait a minute, don't any of those work with (0,x,x)? Whateve, 25 was what you were looking for.)

And I am far too lazy to hunt through your ISO.

[Who]

Actually I forgot that I could just search for "mistake" to limit how many posts I had to hunt through:

The provider of easy math problems to the rescue!

A student was asked to solve sqrt(x + 16) for x = 4 and gave the incorrect answer of 6. Find the mistake the student made, disprove the mistake's validity, and correctly solve sqrt(x + 16) for x = 4 in simplest form.

I'm guessing that that was the mistake, and this is the disproof:

The provider of easy math problems to the rescue!

A student was asked to solve sqrt(x + 16) for x = 4 and gave the incorrect answer of 6. Find the mistake the student made, disprove the mistake's validity, and correctly solve sqrt(x + 16) for x = 4 in simplest form.

The student incorrectly assumed that √(a+b)=√(a)+√(b), which is incorrect for a,b>0.
Proof:
Create a right triangle of leg lengths √a and √b.
√a+√b>√(a+b) because the sum of the lengths of two of the sides is always greater than the third.

√(4+16)=√(20)=√4*√5=2√5

Not sure how 16 follows without using explosion or something similar though.

[Who]

Assume that √(a+b)=√(a)=√(b)
And 4=2+2 thus √(16)=√(4)+√(4) thus 16=4+4 thus 16=2²+2²

This seems a bit obviously wrong though.

[Who]

I meant that it's obviously wrong because 4+4=8.

Also because one could use that logic to show that any square greater than 1 is the sum of two other squares, it doesn't exclusively produce 16.

[Who]

What domain is "arbitrary number"?
Natural, rational, real, complex, something even more vague, ???

[Who]

Sufficiency:
By the properties of logarithms:
If N=1
Log(a)+log(1)=log(a)
Thus log(1)=0.
If N>1:
By FTArith, all natural numbers greater than 1 either are prime or are the product of primes.
For any composite number N, to find logN just sum together the logs of its prime factors.

Necessity:
If you do not know the log of all prime numbers then you do not know the log of all numbers thus for any arbitrary natural number you might not know the log.
Is that what you meant?

[Who]

A new one:
Prove or disprove:
X^2-3 can never be a multiple of 7. (For any integer x)

[Who]

Huh O.o? I don't understand..

First of all, what is a Pappus graph and what do you when you say that the infinite chessboard must "represent" it?

Reposting un-answered puzzle from last page:

On an island there are 1000 people and they all have brown eyes.
They are trapped on the island and can only leave once they know what color their eyes are.
All of the 1000 people can see that everyone else has brown eyes, however, because island is cursed, there is no way for them to easily tell what colour their own eyes are!
The mirrors and the water doesn't reflect light and they can't communicate about eachothers eyecolour.
Once each day, a ship arrives to the island and the Captain tells them that at least 1 of the villagers have brown eyes.
If they have figured out, what colour their own eyecolour are, they will leave with his ship.

The villagers are all logicians with a really high IQ and they know that everyone else is a logician with a really high IQ, so if there is a way for them to know what colour eyes they have, they will surely leave the island!

Does the villagers ever leave the island? How do they do it? and how long does it take?

You also need to assume/state that they all want to leave. The first version I heard had knowing their own eye color be banned with punishment being suicide, and I still say that none of them would attempt to figure out their own eye color that way.

I've also already seen it and will leave it for someone who doesn't already know the answer.

[Who]

It's rather famous. Though most of the time there are m with blue eyes and n with brown eyes, and it is known that everyone has only either blue or brown eyes. But same thing.

And leaving if they know isn't the same as wanting to leave.

[Who]

If we're talking about the screwedupness of infinite sums, why not mention that theorem which says that conditionally convergent series can sum to anything if you just rearrange them.

And those are convergent series, not divergent ones you're assigning a value to.

[Who]

Are we allowed to use computers to find them?

[Who]

[9^(9^(9^(9^(99999999999999999999999)))),9^(9^(9^(9^(99999999999999999999999))))+999,999]
I'm probably not wrong.

[Who]

Reviving the thread with a new(ish) problem:
There are 3 gods, True, False, and Random. (This is different to the version which has been asked several times) True always speaks the truth, False always lies, and what Random does depends on the variation, of which this problem includes several. You don't know which god is which, you only know them as A, B, and C. They only answer yes-or-no questions, and they only do so in their own language, in which "Ja" and "Da" mean yes and no, but you don't know which one's which. In addition, if you ask a god a question they can't answer (So for example, if you ask True a question such that no matter what answer he gives, he's lying), the god's head explodes. Exploding a god's head is allowed, but you can't ask a god whose head as exploded any further questions.
Determine with proof, the minimum number of questions needed to figure out the identities of all the gods, and give questions which would allow you to figure them out, under the following versions of Random:
1. Random either always tells the truth or always lies, which one he does is determined at the beginning.
2. Random always either tells the truth or lies depending on how he's feeling at the time, but if possible he tries to prevent his head from exploding.
3. Random always either tells the truth or lies depending on how he's feeling at the time and will decide before he evaluates your question, and sticks to his decision even if it makes his head explode.
4. Random randomly says "Ja" or "Da" unrelated to your question but his head never explodes
5. Random randomly says "Ja", says "Da", or explodes his head unrelated to your question.

(Several of these have the same/very similar minimalist solution(s).)

[Who]

Less new than you imagined, I think.

Anyway, the puzzle:

There are three gods, A, B, and C. In no particular order, one always tells the truth, one always lies, and one answers randomly. You are allowed to ask three yes-no questions of the gods (each question adressed to a specific god). However, while each god understands English, they will only answer with "Da' or "Ja"- one of which means yes, and one of which means no, but you don't know which is which. What three yes-no questions will allow you to identify which god is which?

Exploding heads changes things. You can do at least some of the versions in mine in only 2 questions.

[Who]

They only answer yes-or-no questions, and they only do so in their own language, in which "Ja" and "Da" mean yes and no, but you don't know which one's which.

Say what O.o
I quess the trick, to figuring out who is who, is to make their heads explode then...

I'll start with the first variation where Random either always tells the truth or always lies, and where which one he does tell is determined at the beginning.
- First I would ask all 3 of them "Are you Truth?". They will answer either Da or Ja
- Then I would ask all 3 of them, if what they had just answered to the first question means Yes. Truth will answer the same as before and both Random and False will answer differently no matter what.

Now I know who is Truth and whether Da means Yes or No in Truth's native tongue.
- I will then ask Truth: "Is this Random?" (while pointing on one of the other 2). Truth will give me the answer and I will then know who all of them are.

I did it in 7 or 3 questions, depending on how you count

That works, but I asked for a minimal solution. You took 7 questions when it can be done in 2.

[Who]

Yes they do.
I don't think it matters though.

[Who]

Another, similar puzzle:
There are N pirates, and they have no gold coins. Angry about this, they decide to stage a mutiny. They go through the pirates in order of rank, highest first, and vote to either lynch that pirate or end the mutiny. (Exact same process as the previous problem, but with no gold coins).
Each pirate has the following goals:
1. Survive
2. If they can survive, they want to lynch as many other pirates as possible.
(Exact same goals as the previous problem, but without the gold coins)

The pirates do not trust each other, and will not make deals.
All pirates are perfectly logical.
All pirates know all of this listed information, including this.

How many pirates survive the mutiny?

[Who]

You're outside a small shack with no windows or cracks in the walls and a single door, all very well sealed so no light or air escapes. Inside the shack is a halogen light bulb. Outside the shack are three light switches. Only one of the switches controls the light. You may flip the switches as many times as you like, but once you open the door you cannot flip any more switches. How do you determine which switch is connected to the light with 100% accuracy?

Are creative solutions allowed? (For example, get your friend to flip a light switch after you open the door, setting up some sort of delay mechanism, etc.)

If not, it's impossible. There are only 2 ending possibilities (Light on, light off) but 3 solution possibilities.

[Who]

Is the second hand discrete or continuous?

[Who]

Take 1000 apples 1000/3 miles towards Rome x3, take 1000 apples 500 miles towards rome x2, take 1000 apples the final 1000/6 miles towards Rome, net amount transported: 5000/6.

[Who]

I see no question.

[Who]

edited smartypants

I was genuinely confused, not being a smartass. Of course, now is the time for smartassery: Still no question.

B)
It's obviously true for n=1.
Suppose it's true for n=k.
For it to be not true for n=k+1, that is, for a subset of cardinality n+1 in which such a pair does not occur, it must contain both 2n and 2n-1. (Otherwise, one could just remove a number and arrive at a counter example to n=k, which we assumed).
But 2n and 2n-1 are mutually prime.
Thus no such subset can exist for k+1.
This by weak induction, it's true for all n.

[Who]

Any subset of the form (all the odds +1 even) must have 2 numbers which one divides the other, the even and the odd largest odd to divide that even.
To try and get the subset we want (One without any numbers which divide each other) we take away the largest odd to divide that even and replace it with another even.
But the problem with that is that either the new even has the same largest odd which divides it (In which case it is either divides or is divided by the first even) or is divided by a different odd. Which doesn't work either.
And then we can continue to replace odds using the latter option until our only option is the former at which point we give up and declare that no such subset exists.
(A more rigorous version of this would use induction on number of evens in the subset, but I'm too lazy to write that out)

[Who]

Hint: You're on the right track with the cipher itself.

Hint: Stop trying to come up with keys for a two-square cipher. That's not what I used.

*Walks away*

It's probably a two-square or four-square cipher.

So, four-square?

[Who]

Using this, we can find the x value that produces any desired y-value, so let's do that.

You assume that the "infinite tower" function is surjective onto the reals in this step. It isn't. There is no x-value which gives y=4.

In less mathy terms, no you can't produce any desired y-value.

[Who]

Sure.

Assume it is surjective. Your proof works. 2=4. Contradiction.

assume you actually mean "find the actual domain and range"?

[Who]

If x>e^(1/e) then the sequence defining IPT increases without bound. Proof:
Suppose it is bounded above. Then it has a supremum. x^x is C1 and increasing. x^x-X is also C1, increasing, and positive when X>e^1/e. Thus, this supremum must be greater than the original X. Call the supremum s. s^s>s, but because it's continuous we can find some point epsilon below s such that (s-\ep)^(s-\ep)>s. Because s is the supremum, we can find a point in the sequence which is within epsilon of s. But then the next point in the sequence is greater than s. Contradiction.

[Who]

Thus the domain is [0,e^1/e] or some subset of that. Range time.

[Who]

If x\leq 1, x^x\leq 1, thus y \leq 1.
If 1<X<e^1/e, x^x is increasing, thus y is increasing, thus y\leq y(e^1/e)

y(e^1/e)=e by the formula you gave. Thus no y>e.

[Who]

True. Also possibly other things if we allow complex y values.

[Who]

The function x^x-x was referenced to get when the sequence was increasing, that is, when x^x is greater than X.

[Who]

Wait you're right. Nevermind.

[Who]

My sequence is a subsequence of yours.

[Who]

Wait I didn't need the fancy analysis.

X^1/X has a maximum at e. Thus, no number greater than e^1/e is expressable as x^1/X, thus no y exists.

[Who]

Could you put that in less mathy terms?

[Who]

Split them into 2 piles of 10. There are x heads in pile 1, and (10-x) heads in pile 2. Now flip all of the coins in pile 2. There are now (10-(10-x)) in pile 2, which is the same as x. Thus, there are the same number of heads in pile 1 as pile 2. (And by extension the same number of tails)

[Who]

Impossible.

If I hear the number "1", there are three options:
Guess "you told me the greater number" with a <50% chance
Guess either with equal chance.
Guess "you told me the lower number" with a <50% chance

You know my strategy. Thus you know which option I will take. If I take the first option, you pick 1 and 0, I have <50% chance of winning. If I take the second option, I lose. If I take the third option, you pick 1 and 2, I have a <50% chance of winning.

Thus it's impossible to create a strategy which survives you knowing the strategy. Probably impossible even without that but definitely impossible with.

[Who]

If the chooser knows that in advance they can counter it.

Take probability that c ends up in (-infty,1) and probability that c ends up in (1,infty) and see which is bigger. Tell you 1 and a number greater/less than that with the higher probability of you losing.

[Who]

Another way of looking at it without relying on knowing your strategy:
You have to beat every single chooser. One of those choosers will tell you always 1 and have the other number be 0. Another will always tell you 1 and have the other number be 2. If you can beat both of those choosers, you are doing something very wrong.

[Who]

Proof of at least one:
All elementary column operations are the result of multiplying by an elementary matrix. Thus, column reduce a matrix. There are two options:
You run into a column of all zeros, one column was a linear multiple of the others.
You get the identity, multiply those elementary matrices together to find the inverse.

[Who]

Proof of exactly one:
Suppose you have a column being linear multiple of the others. Perform elementary column operations until you get it to be a column of all zeros. That new matrix can't have an inverse because it has a column of zeros. But the elementary matrices you multiplied by do have an inverse, so the original matrix couldn't have an inverse either.

[Who]

Do they arrive sequentially or at the same time?

[Who]

They're the same size.

There exists a bijection from whole numbers greater than zero to rational numbers between 0 and 1 (Cantor), there exists a bijection from rational numbers between 0 and 1 to rational numbers between 3 and 4 (x+3), and any infinite subset of a countably infinite set is the same size as the original set.

[Who]

My answer assumed cardinals, ignore it.

[Who]

There are limits of sets. There are limsups and liminfs (Union of intersections and intersection of unions, I forget which one's which, it's relatively easy to figure out), and if limsup=liminf, that's the limit, similar to limits in the reals.

Edit: Liminf is union of intersections. It contains everything eventually in the sets. Limsup is intersection of unions, it contains everything which is in the sets infinitely often.

[Who]

Am I doing it wrong if I got a system of 7 equations 6 variables? It feels like there should be an easier way. (The equations look annoying to solve by hand and they're in base 6 so they'd be annoying to enter into a computer)

[Who]

Spoiler: When in doubt add more variables

x_n is the amount of beer each man starts with.
y_n is the amount of beer each man gives to each other man. (So 1/6 of what each person has when they pour)
The 7th man has no beer. x_7=0.
The 6th man has only what the 7th man gave him. x_6 = y_7=y_7+x_7
The 5th man has only what the 7th and 6th men gave him. x_5 = y_6+y_7= y_6+x_6
The 4th man has only what the 7th, 6th, and 5th men gave him. x_4 = y_5+y_6+y_7 = y_5+y_6+x_6 = y_5+x_5
Blah blah blah induction, x_n=y_{n+1}+x_{n+1}
y_n?
y_1=x_1/6 = (y_2+x_2)/6 = (y_2+y_3+x_3)/6 = ...=(y_2+...+y_7)/6
y_2=(y_1+x_2)/6 = ...= (y_1+y_3+...+y_7)/6
...
y_7=(y_1+...+y_6)/6
So basically 6y_n is the sum of all the other y_ns.
Thus 7y_n is the sum of all the y_ns.
Thus all the y_ns are the same.
Thus they all have the same amount when they pour.
Since they're all the same let's just call it "y".
x_n=y+x_{n+1}
So x_6=y, x_5=2y, x_4=3y, and so on.
So we end up with y+2y+3y+4y+5y+6y=42
y=2
So x_6=2, x_5=4,x_4=6, x_3 = 8, x_2 = 10, x_1 = 12

[Who]

Suppose there's a room that contains at least two martians, each martian has at least one finger. You take a quick scan, and note that there are between 200 and 300 martian fingers in the room, and that each martian has the same number of fingers. From this knowledge, you then know exactly how many martians are in the room. So, how many martians are in the room, and how many fingers does each martian have?

Do you start with the knowledge of how many fingers each Martian has?

[Who]

It still seems unsolvable, unless I'm misreading it somehow.

There are m martians and f fingers per martian.
m\geq2
f\geq 2
200\leq mf \leq 300

There are lots and lots of solutions. Unless the scanner has access to information the reader doesn't, it's impossible for them to figure out what they are said to know.

[Who]

The second question to this is actually pretty different from the first, in that I think FakeGod's method won't work. Think about something like Markov chains.

A video game requires you to slay monsters to collect gems. Every time you slay a monster, it drops one of three types of gems: a common gem, an uncommon gem or a rare gem. The probabilities of these gems being dropped are in the ratio of 3:2:1 — three common gems for every two uncommon gems for every one rare gem, on average. If you slay monsters until you have at least one of each of the three types of gems, how many of the most common gems will you end up with, on average?

An additional question for the monster slaying riddle:

what is the expected number of monsters you need to slay to have an odd number of each gem?

Are we allowed to use a calculator to do the annoying matrix inversion/multiplication?

[Who]

Call the elements of C x and y. We know that xy=105-x-y. x+xy+y=105. I forget how to solve equations of that form so let's use brute force. 105=0 mod 7. Thus xy=(7-x)-y.
If x=1:
y=6-y, y=3.
2: 2y=5-y, y=4
3 and 4 via symmetry
6 is impossible, 6y=1-y, 7y=1, 0=1
Thus 5 must pair with 5
Thus the only pairs we need look at are:
(1,3),(1,10),(8,3),(8,10)
(2,4),(2,11),(9,4),(9,11)
(5,5),(5,12),(12,12)

Elimination time:
1,3 and 1,10 are obviously wrong.
8+24+3 nope
8+80+10 nope
2,4 and 2,11 obviously wrong
9+36+4 nope
9+99+11 nope
5,5 can't repeat
5+60+12 nope
12,12 can't repeat.

It is impossible.

[Who]

Oh oops it's +1 not -1.
Redone using my method:

Spoiler:

The pairs are now (0,1), (2,2),(3,3),(4,5),(5,4),(6,nope)
0,1: (0,1),(0,8),(7,1),(7,8)
2,2: (2,9)
3,3: (3,10)
4,5: (4,5),(4,12),(11,5),(11,12)

Testing:
0,1 nope
0,8 nope
7,1 nope
7+56+8 = 71 nope
2,9 nope
3,10 nope
4,5 nope
4+48+12=64 nope
11+55+5=71 nope
11+132+12 = nope
No possible solutions.

[Who]

It starts with the bottom and bottom left bars broken. Then the top left bar breaks just as the 9 changes to 0.

Edit: I don't know how displays work and thus don't know if this answer makes sense: The top left bar needs to be a part of a loop because the clock designers were lazy and figured it would only be on when it was part of a loop. So when it's part of a 9 it's fine but when it's part of a 0 the other two broken bars make it also broken.

[Who]

Any one made any progress on implosion's problem from a while ago, by the way?

I'm pretty sure it's player 1 wins unless (n-1) is a multiple of 3. Working on a proof.
EDIT: My other method was wrong. I don't know if the answer's right though.

[Who]

Spoiler: Implosion's problem

Rephrasing it to subtract one from everything then allow the last stone to be taken because that seems easier:
n=0:
Player 1 loses
n=1
Player 1 takes 1 stone, player 2 loses.
n=2
Player 1 takes 1 stone from the pile with 2, player 2 loses.
n=3
Player 1 takes 2 stones from the pile with 3, player 1 loses.
n=4
Player 1 takes 1 stone from the pile with 4, player 2 takes 2 stones from any possible pile, player 1 takes 3 stones from a pile with 3 (Must be at least one left), player 2 loses.
n=5
Player 1 takes 1 from 4, (53321) player 2 takes 2 from anything, player 1 takes 3 from 5, no 4s left, player 2 loses.
n=6 (654321)
Player 1 wants to either take away all the 4s or preserve a 5. If P1 takes from the 6, P2 takes from the 5. Otherwise P2 takes from the 6. There are now 2 4s, 1 5. (When I'm referring to them like this I mean "at least", so two piles have at least 4, one has at least 5. These overlap). P1 can destroy one of them or not it doesn't matter, P2 has a 4 left and destroys the 5 if it still exist, P1 has no 5s, P1 loses.
n=7 (7654321)
Player 1 has established that he can't take away all the 4s, but maybe this time he can preserve a 5 while taking away all the 6s. Take from the 6, there are 3 5s 1 6 left. P2 can only destroy 1 5 per turn, P1 will have a 5 left and can destroy the remaining 6 when he takes his 5.
n=8 (87654321)
Player 1 takes away from the 6 (8754321). On his next turns he destroys the two remaining 6s, player 2 loses.
n=9 (987654321)
Player 1 can't destroy all the 6s, player 2 can destroy all the 7s. I see a pattern here.
n=10 (10987654321)
Player 1 can destroy all the 8s, player 2 can't destroy all the 7s.
According to implosion's phrasing, player 1 wins when n-1 is not a multiple of 3.

Proof will use my phrasing:
The game ends on turn n+1-floor((n+1)/3). Hmm okay maybe this proof would be neater with Implosion's phrasing. Oh well, we're using mine.
Proof:
Player 1 can destroy x moves by turn 2x-1. Player 2 can destroy x moves by turn 2x. There are n+1-k moves to be destroyed before the kth turn cannot be played. (And thus, if it is the first such turn, the turn player loses).
Thus, player 1 can only destroy a turn if 2(n+1-k)-1 < k.
2n+2-2k-1<k
2n+1<3k
k>(2n+1)/3
Player 2 can only destroy a turn if 2(n+1-k)<k
2n+2-2k<k
k>(2n+2)/3

If n=3m (n is a multiple of 3):
The earliest turn player 1 can destroy is (6m+1)/3, which is 2m+1/3, so earliest turn he can destroy is turn 2m+1, which is odd, earliest of p2's he can destroy is 2m+2.
The earliest turn player 2 can destroy is (6m+2)/3, which is 2m+2/3, so earliest turn he can destroy is turn 2m+1, which is odd, he wins.
If n=3m+1:
The earliest turn player 1 can destroy is (6m+3)/3, which is 2m+1, so earliest turn he can destroy is turn 2m+1, which is odd, earliest of p2's he can destroy is 2m+2.
The earliest turn player 2 can destroy is (6m+4)/3, which is 2m+1+1/3, so earliest turn he can destroy is turn 2m+2, he loses.
If n=3m+2:
The earliest turn player 1 can destroy is (6m+5)/3, which is 2m+1+1/3, so earliest turn he can destroy is turn 2m+2, which is even.
The earliest turn player 2 can destroy is (6m+6)/3, which is 2m+2, so earliest turn he can destroy is turn 2m+2, he loses.

The proof is incomplete, it still needs to be shown why when you need to destroy your own moves that doesn't make a difference, but that's basically why.

[Who]

Spoiler: solution

Indeed, there is no winning strategy. If you arrange the numbers into a magic square, like:

8 1 6
3 5 7
4 9 2

Then it becomes clear that the game is functionally identical to Tic Tac Toe, which is always a draw with optimal play

The problem said "whose claimed tiles sum to 15", nothing about containing a subset which sums to 15. Tic Tac Toe allows you to have pieces on the board not in the winning 3. Thus, you can threaten multiple attacks at once, and you have to look more than 1 move ahead in order to force a draw. Furthermore, not all sums have only 3 pieces. For example: 1,2,3,9 would win but not on that magic square tic tac toe board.

[Who]

Do you need to know that they're all facing the same direction to win?

[Who]

Reach in to diagonals, flip both up, UXUX.
Reach into adjacent, flip both down, DDUX.
Reach into diagonals, if one is up and one is down, flip both to opposite. UDDX OR DUUD. If both are down, you have DDUD, flip one of them, you now have either DDUU or DUUD.
Reach into adjacent. If both are up flip both down you're done. If both are down you might be at UDDD from earlier, or you might be at UUDD. Flip both up, now you're either done or at UUUD. If one up one down, flip both then reach into diagonals and flip both then you're done.
If you've gotten to this point you must be at UUUD. Reach into diagonals. If they're different flip the down one up, you're done. If they're the same flip one down, now you're at UUDD.
Reach into adjacent, if same flip both you are done, if different flip both then
Reach into diagonals, flip both, you are done.

7 moves is best I could do.

[Who]

Wait, I can do better.
Reach into adjacent, flip both up, UUXX.
Reach into diagonals, flip both up, UUUX. If you haven't won you must be at UUUD.
Reach into diagonals. If they're different flip the down one up, you're done. If they're the same flip one down, now you're at UUDD.
Reach into adjacent, if same flip both you are done, if different flip both then
Reach into diagonals, flip both, you are done.

5 moves.

(My first solution didn't fully take advantage of the fact that you'd be told you were done, so it never made a purely information move like the second move in this one)

[Who]

What do you mean by "bend up in the middle"?
Is it an arc, a triangle, some other fancy shape?

[Who]

Spoiler:

Qd4.
If black blocks with pawn, Qd5#. Else f7#.

[Who]

Personally I prefer other things to chess puzzles but other things aren't happening so go ahead and flood the thread.

[Who]

Timmy handed him four quarters, Tommy handed him one dollar bill.

(Timmy may also have used some other smaller coin but who uses those?)

[Who]

Ooh here's one I composed a few years ago. Mate in 5.

Spoiler:

Bd8 b4 Bg5 f4 Bd8 f3 Bg5 e4 Bf6#
You're always threatening mate in one move if they deviate from that. (Other than moves symmetric to that, which lead to essentially the same game)

[Who]

Here's one for the game theorists out there:

You're playing Now or Then on

The Price Is Right

. Six grocery items are arranged in a circle, each with a price displayed. You must select an item. If you think the price displayed is current, you must guess NOW. If you think the price displayed is from the past month and year you're told at the beginning of the game, you must guess THEN. You are told whether you're right or wrong after each guess, and the prices may be attempted in any order. If you guess correctly on three adjacent items in the circle, you win the pricing game and a prize; you lose when the win condition becomes impossible. Although the host doesn't tell you this, you know from analyzing past playings of the game that the distribution of the NOW and THEN prices is fixed at four NOWs and two THENs rather than determined randomly.*

Prove that you can force a win despite knowing nothing about past or present prices of any of the items and describe how.

* That's true of the actual pricing game on the U.S. version of the show, by the way.

Spoiler:

Label them so they are arranged 123456 (With 6 adjacent to 1):
Guess NOW on 1.
If wrong:
Gues NOW on 6,5,3,2 in that order. If you have hit a THEN in one of those, guess NOW on 4 to win. If you haven't, guess THEN on 4 to win.
If right:
Guess NOW on 2. If you're wrong, you're basically at wrong on 1 except you've already guessed and been right about 6, so just complete that strategy.
If you're right, guess NOW on 3. If you're right you win. If wrong you're basically at wrong on 1 except you've already guessed and been right about 6 and 5 so just complete that strategy.

[Who]

Spoiler:

The probability that mean < .1 is the probability that sum < .2, so:
y+x<.2
y<.2-x
Geometrically, the whole space is a 1x1 square, this is a triangle in the bottom left corner, with vertices (.2,0) (0,.2) and (0,0). Total area of this is .2^2/2 = .02
Thus the probability is .02.
(< and <= are same by nonatomicness)

[Who]

Spoiler: 1)

For any fraction x, find an n such that 1/n<x. Divide the prisoners into groups of n, use standard prisoners and hats solution for each of those groups. (First prisoner out of the n says black if even number of black hats in his group, white if odd, then next prisoner knows what his hat is from that, etc.)

[Who]

I'm pretty sure that part 2 is impossible. Working on a proof.

The proof isn't working, meaning it's probably possible, or at least, if it's impossible it's for a different reason than I was trying to use.

[Who]

Wait what? What do you mean "can't hear previous responses"? Can't hear the guesses? Then it's definitely impossible.

[Who]

Spoiler:

Let S be the set of functions which map binary strings to the set {0,1}.
By the well ordering theorem, S has a well-ordering. Agree on such a well ordering in advance.
Each prisoner looks at the prisoners in front of him, looks at the elements of S, finds all functions which would lead to only a finite number of prisoners dying, picks the least such function, and applies it to himself.

They will all choose the same function. Proof:
If prisoner m sees a finite number of prisoners live with the function, prisoner n will too. There are only a finite number in between them. Thus, the sets both are dealing with are the same, thus both sets will have the same least element, thus they'll both pick the same function.

There is always such a function. Proof:
There are two cases:
If the string does not eventually repeat forever, no substring matches another substring which starts at a different place. Thus, there exists a function which saves everyone.
If the string does eventually repeat, there is a function which saves everyone in the repeating part. (Just guess the hat of the thing in the repeat before the next part of it you see), thus only the prisoners before the repeat die, thus there exists a function in which finitely many prisoners die.
Thus the subset is nonempty.

[Who]

For part b, do they get to hear whether or not the person behind them lived?

Irrelevant.

Spoiler: a

The original solution but replace the 2s with ks.

Spoiler: b

The first prisoner checks the function, sees the last prisoner who would have died, says the parity up to that prisoner. (Might die). If nobody ahead of him dies guesses according to the function.
The nth prisoner checks the function found, if anyone ahead of him would die by it he guesses according to parity and previous guesses. If nobody ahead would die, guesses according to the function.
(Prisoner who would have died last by the function will die, because since nobody ahead of him dies he guesses according to the function)
At most 2 deaths. I don't think it's possible to get at most 1 death, but no proof.

[Who]

What is 3+2?

10.
(In base 10)

[Who]

Spoiler: Gold

1) Weigh 1 ingot from the first vassal, 2 from the 2nd, 3 from the 3rd, 4 from the 4th, none from the 5th.
If the total is 100 kg, it's the 5th vassal. If the total is 104 kg or 96 kg, it's the 4th. If it's 103 or 97, it's the 3rd. If it's 102 or 98, it's the 2nd, if it's 99 or 101, it's the first.
2) 1,2,4,8,16. Subtract the weight from 310, convert to binary, look at digits.
3 I need to think about more in case there's a way which involves touching less gold, I don't want to be put to death. My gut instinct says 1 3 6 9 27 81, but that seems inefficient. If he had more vassals I know of tricks which might help, but he has too few vassals for them.

[Who]

2 proof)
There are 32 total possible combinations of fakes. If n ingots are weighed, the scale will return an integer result from 9n to 10n, inclusive. Thus, there are n+1 total possible answers the scale can give. There must be a surjection from the set of scale results to possibilities that we use to determine guilt, thus the set of answers must be at least as big as the set of possibilities, thus the set of answers must be at least of size 32, thus there must be at least 31 ingots used.
1 proof:
The amount from two different vassals cannot be the same, if it were the same then it would be impossible to distinguish which one is guilty of exactly one of those two is guilty. All ingot amounts must be nonnegative, thus it cannot decrease below 0, all must be integers, I took the set of 5 distinct nonnegative integers with the smallest sum.
3 is in fact 1 3 9 27 81. Proof incoming.

[Who]

The proof I thought would work for 3 is blatantly wrong. (I thought that it would just be an argument about how many possibilities like 2, but then with the added section where you prove that two different but equivalent possibilities (For example: vassal 1 gives 9 and vassal 1 gives 11) can't cancel out, but that line of argument won't work because there are 364 total possibilities for the scale to return with 121 ingots, but there are only 243 possibilities for answers, that argument just says you need at least 81 ingots) Also I'm back to thinking that the answer is wrong.

I cannot do basic arithmetic.

[Who]

Spoiler: Proof of 3

Lemma: For any two disjoint nonempty sets of vassals, the total number of ingots on the scale from the first set cannot equal the total number of ingots on the scale from the second set. If this were the case, it would be impossible to distinguish between (first set is guilty with all 9s and second set is innocent) and (second set is guilty with all 9s and first set is innocent).
The scale can return anything from 9n to 11n, thus there are 2n+1 possible answers. There are 3^5=243 possibilities, thus we need at least 121 ingots if we are to detect all possibilities.
"But wait!" says the King, "I don't need you to distinguish between all possibilities, only guilt and innocence. I don't care if you know that Vassal 1 gave me 9kg ingots or 11, I only care that you know it's fake".
The King is correct, but unfortunately it's impossible to map equivalent strings to the same weight if we are to preserve the property that nonequivalent strings map to different weights. Proof:
Suppose we have numbers of ingots on the scale which map two equivalent strings to the same weight. That is, you can replace some vassals who gave 9 with 11 and some vassals who gave 11 with 9, and the total weight will be the same. (You do not replace any 10s, and you do not replace anything with 10). Then, the total ingots provided by the vassals who gave 9 which got swapped to 11 is the same as the total ingots provided by the vassals who gave 11 which got swapped to 9. This is illegal, as per the lemma, unless both sets are empty. (If one's empty the other has to be too, since everyone needs an ingot on there to distinguish between everyone real and everyone but them real)
Thus, the type of fake must be distinguished, thus all possibilities need to be accounted for, thus at least 121 ingots are needed.

Also, since I don't think I ever actually said: To find guilt just subtract 9*121 from the total returned, convert to base 3, anyone who has a 1 in their corresponding digits place is legit, anyone with a 0 or a 2 is fake.

Spoiler: Trick I wanted to use but would not have returned enough information even if it would beat 3^n

If there were more vassals, I could in binary add an unbounded constant number of bits, or add a logarithmic number of bits, or add bits and have the last added bit be not a bit but a a value. What I wanted to do was add 32 ingots from every vassal, thereby telling the sum of the total offness, I thought that from there I might have been able to do something, but I couldn't because the proof generalizes to any number of vassals. Also for a while I considered adding a bit to every pair of bits, yielding which I think might lose to trinary with 5 vassals but not more. None of these would have worked, the proof generalizes to any positive integer number of vassals.

[Who]

I think it's RRB but I can't come up with a proof of optimality. Brute force seems possible, it's only 7 possibilities, but that feels like cheating.
General solution would be (not2)(1)(2), so that to get theirs they have to not be yours (Yours ends in 12, meaning that their 3 card trick 123 is secretly four cards, 2123, unless it's the first three cards), and by picking not-2 they don't do the same thing to you. (BRB would steal their tricks but also make them steal yours)
Also I treated it as random binary strings with 50% chance of either being next, I don't think the fact that it's cards changes that much but I'm not 100% sure.

[Who]

Is it assumed that all goods (Lamb/Harmonica) are worth an integer number of dollars?

Given that there are x cows it has to be

[Who]

Real answer: Neither is correct. It doesn't make sense to talk about who brought "more" because they both brought a divergent sum. What does make sense, is to say that opening envelopes takes time, assuming that you can't just open an arbitrary envelope and either you have to open all the envelopes which precede any envelope you open or you have to at least flip through the envelopes until you find the one you want (So opening the nth envelope without the preceding ones takes linear time based on n), the second messenger brought money which was easier to access, and thus you should aid him.
Specific refutations to both arguments:
They both brought the same number of envelopes, so the first one's claim that the second only brought half of his is wrong.
The claim of the second is true, if you look at it in a specific way. But there are different orderings of envelopes which lead to most of the first's being greater. For example, if you reorder the envelopes of the first to go 1,2,3,4,8,16,32,5,64,128,256,512,1024,2048,6,... (2^n except every 2^n elements there's a small element), yes the first occasionally has an envelope smaller than the second but after every envelope opened after the first few the first has given much much more.

Silly answer: Uhh, "bribes" should be positive amounts of money, not negative. The first one gave -1/12$. The second one gave -1/6$. (That's how it works, right?) Send both of these con artists home. If you're forced to aid one aid the first one I guess because he took less money from you.

[Who]

In 1), are the disks labeled by the natural numbers? Also what do you mean by "proportion"? The limit of the proportions of arbitrarily large but finite ToHs? There are the same number of valid configurations as there are configurations.

[Who]

I think you're much better off not using the word "infinite".

An infinite tower of Hanoi could exist but it would look very different to the arbitrarily large ones you're talking about. There are continuum many configurations and continuum many valid configurations, so proportions make no sense. "Average" doesn't make much sense unless they're all (or almost all) the same.

[Who]

The sets are both countably infinite

No, they're uncountable.

Consider the function f maps valid configurations to the power set of N, where f(configuration) = the set of disks on the first peg. f is surjective, thus the sets are at least continuum. (They are continuum, not larger. There are two proofs of this, you could either be formal and define an injection to P(N)^3 times [the set of functions from N to N]^3, or you could just say its obvious because they can't be that big)

Maximum number of steps is 2^n-1, the number of steps needed to solve the game in its normal version.

[Who]

It's not obvious that they're the same until you try to prove that either of them is 2^n-1. Simple induction yields either result.

[Who]

Save the treat it as heads for coin 9 or 10. Use it on 9 if 9 is tails, use it on 10 if 9 isn't and 10 is. The expected loss is 41$.

Work:
Expected loss for any given coin with no treat it as heads is i/2+i/4+...=i. Expected loss for any given coin when it just came up tails and deciding whether or not to treat it as heads is 2i (Because you just got tails and then you're back to the beginning). So if you flip tails at 9, expected gain from treating it as heads is 18, expected loss is 10. If you flip tails at 8, expected gain from treating it as heads is 16, expected loss is 19. Your total expected loss is 1+2+...+8=36, plus .5*(Expected loss of coin 10) (If 9 is heads then 10 is automatic heads, if 9 is tails you lose the expected loss of 10)

[Who]

Expected Value = f(x) = Summation(i / 2 ^ x).

As x --> +Inf, f(x) --> 2i. (I just caslculated the approx value from 0 to 20 then rounded it fyi)

Therefore, total EV = $2 +$4 + $6 +$8 + $10 + $12 + $14 + $16 + $18 + $20 = $110

Since the tenth coin obviously has the highest EV, save the count as head for that coin giving an EV of 0 for that coin (unless I misread the prompt). Therefore, expected loss equals $110 - $20 or $90.

You don't have to pay if it comes up heads the first time, meaning i starts at 1.

[Who]

Save the treat it as heads for coin 9 or 10. Use it on 9 if 9 is tails, use it on 10 if 9 isn't and 10 is. The expected loss is 41$.

Work:
Expected loss for any given coin with no treat it as heads is i/2+i/4+...=i. Expected loss for any given coin when it just came up tails and deciding whether or not to treat it as heads is 2i (Because you just got tails and then you're back to the beginning). So if you flip tails at 9, expected gain from treating it as heads is 18, expected loss is 10. If you flip tails at 8, expected gain from treating it as heads is 16, expected loss is 19. Your total expected loss is 1+2+...+8=36, plus .5*(Expected loss of coin 10) (If 9 is heads then 10 is automatic heads, if 9 is tails you lose the expected loss of 10)

My elimination of 8 was wrong, I don't know what I was thinking when I said it.

Expected loss from using is 19-5=14, expected gain is 16. Use it on 8. New expected loss of end is .5*5+.5*19=12. Expected loss entering 8 with nothing is 10+9+8=27, making the amount saved 17, which is greater than 14, expected gain from using it at 7.

Actual best strategy is to save it for 8 then use it at first opportunity. Expected loss is 28+12=40.

[Who]

I have a puzzle for you all. It's described here:

https://en.wikipedia.org/wiki/Hodge_conjecture

Solve it and don't tell anyone other than me and I will give you $5.

Saw the word "algebra", gave up.

A chess puzzle:
Consider infinite chess. That is, there is an infinite (in all directions) grid, and each player has pieces at certain locations in the grid. Both players have exactly one king, but any number (Including infinitely many) other pieces. The game is over when a king is taken, and the player whose king was taken loses. Does there exist a starting position from which white has a winning strategy, but black can delay their loss by as long as they want? (That is, for all natural n, black has a strategy such that white does not win in the first n moves, but for all strategies black can play, white will eventually win)

[Who]

Oh I miswrote that. Both players can have any number (Even infinite) of other pieces.

[Who]

White doesn't have a winning strategy. It's impossible to checkmate with rook and King vs King without an edge of the board.

[Who]

That does indeed work if you ignore the 50 move rule. What about if you don't ignore it?

[Who]

I'm not so sure it's possible to set that up so that white can't can't hide behind the pawns. Could you perhaps provide a more explicit construction?

[Who]

Black can't prevent losing forever, they have to pick some finite number of turns in the beginning (Or at least, in the beginning by most scenarios anyone would come up with unless they want to make things unnecessary complicated) and they'll last that long.

Look at Something_Smart's scenario. If you name any positive integer, black can survive that long. As soon as black moves their castle, white has a winning strategy in a finite number of moves.

[Who]

I feel like I'm reading 4 wrong. Isn't x^3 an easy counterexample?

[Who]

Spoiler: #4

First, it's true in R. This is because any polynomial in R is eventually mostly just its highest power, and is eventually either increasing, or decreasing fast enough to go below zero. Since the latter is impossible, we need only look at the former, thus we only need to look at it over some closed bounded interval, and it must obtain its minimum over that interval.

We can parametrize the two dimensional polynomial into a one-dimensional polynomial in terms of t by taking a line going out from the origin. Thus by the fact that its true in R, all of these lines obtain their minima.
Call the polynomial p. f maps theta to min_{t\in\R}(p(t\cos\theta, t\sin\theta) is continuous. This is because at the t-value for any given min, nearby you'll have a point nearby, making f(nearby thetas) less than or equal to nearby f(theta). Since this applies in both directions, this means that f(nearby thetas) are nearby f(theta). Thus, f has a minimum on [0,2\pi]. This minimum is the minimum of the entire function.

[Who]

Spoiler:

That's not the false claim. If as you say, there's a positive polynomial, then the thing about every line achieving its minimum is false.

[Who]

Spoiler:

I was misreading your counterexample. By "along any line close to it it has no minimum" I thought you meant "for any given line close to it, on that line it has no minimum"

[Who]

3b:

Spoiler:

Ask the following questions to the following guards in order. If they say "Yes", go with the yes:. If they say no, go to the next guard.
1 2 3 4 5 ("lies" means arbitrary, not lies)
1: Is 5 truth?
Yes: Ask 5 about everyone else.
2: Is 1 truth?
Yes: 1 is truth, 5 is lies. Ask 1 if 2 is truth, then 3. 4 must be whatever remains.
3: Is 2 truth?
Yes: 2 is truth, 1 is lies. Ask 2 if 3 is truth, then 4. 5 must be whatever remains.
4: Is 3 truth?
Yes: 3 is truth, 2 is lies. If 1 is truth then 5 must be lies, meaning 1/3/4. If 1 is not truth then 3/4/5.
5: Is 4 truth?
Yes: 4 is truth, 3 is lies, if 1 is truth then 5 must be lies, and so is 2, illegal. Thus it's 4/2/5.
No: If all of them said the previous one was lies, there can't be two truths in a row. Thus it's 1/3/5.

3c:

Spoiler:

The arbitrary guards could answer every question as if they were the truth tellers and the real truth tellers were arbitrary. It would be impossible to distinguish between the two groups.