Math and Logic Puzzles: Redux

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vizIIsto
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Post Post #100  (ISO)  » Tue Oct 29, 2019 7:35 am

In post 99, Mitillos wrote:Lend the dead man a camel.


That still doesn't answer the question how you will divide the seventeen camels among the three sons. Give a complete answer with reasoning.

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Post Post #101  (ISO)  » Tue Oct 29, 2019 12:01 pm

Since you insist: You can't. By the time the will takes effect, the trader is dead. It is impossible to change the number of camels, assuming you have provided all pertinent information, thus at least one camel will have to be chopped up.
If we allow silly solutions like lending a dead man a camel, he now has 18 camels, which can be divided as you have requested. The sum I gave shows that one camel will be left over, which can then be returned to the lender.
You don't have ambiguity; you have options.

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Post Post #102  (ISO)  » Tue Oct 29, 2019 12:12 pm

In post 101, Mitillos wrote:Since you insist: You can't. By the time the will takes effect, the trader is dead. It is impossible to change the number of camels, assuming you have provided all pertinent information, thus at least one camel will have to be chopped up.
If we allow silly solutions like lending a dead man a camel, he now has 18 camels, which can be divided as you have requested. The sum I gave shows that one camel will be left over, which can then be returned to the lender.


And that's exactly how to solve the problem.
1/2 of 18 is 9; 1/3 of 18 is 6; 1/9 of 18 is 2. 9 + 6 + 2 = 17. You don't need to 'lend a camel' to solve it, but adding an extra camel is how you will be able to divide the 17 camels fair and square.

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Post Post #103  (ISO)  » Tue Oct 29, 2019 12:31 pm

the dead man had a problem with adding to 100%

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Post Post #104  (ISO)  » Tue Oct 29, 2019 1:20 pm

In post 102, vizIIsto wrote:You don't need to 'lend a camel' to solve it, but adding an extra camel is how you will be able to divide the 17 camels fair and square.


This is extremely incorrect. The only way to divide the 17 camels "fair and square" is to give each of the sons the amount stipulated in the will, including part-camels.
To see why this is true, consider a similar scenario, where the father leaves one sixth of his camels to each of his three sons and dictates that the remainder are to be released. Let's say he happens to have 18 camels. Normally, this would mean that each son gets 3 camels, with 9 left-over and released. If we accept the idea of adding camels and then removing them, we could add another 18 camels for a total of 36, give each son 6 camels (twice what the will states), and then return the remaining 18 camels to wherever we got them from, leaving no released camels. This is a ridiculous way to do a will, given that it results in going in opposition to what the deceased stated their intent to be.
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Post Post #105  (ISO)  » Tue Oct 29, 2019 9:20 pm

For the ones who like puzzles, solve this sudoku:
9236
651
5678
894
39
457
4593
927
3854

Bonus points if you can prove you've done it manually without the help of a solver, since it's apparently really hard >=) (I created it, so I know the solution, but still...)
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Post Post #106  (ISO)  » Thu Oct 31, 2019 9:58 am

Spoiler: StrangerCoug's sudoku
189273654
367458912
524961738
852739146
736142895
941586327
415697283
698324571
273815469


after filling in the trivial squares, i did some hypotheticals with the 7s by putting my fingers on the paper to show where the slots would be filled, the first place i put my fingers didn't yield anything promising either way, but the second place, the 2nd row 3rd column, if i hypothetically filled that one in, then all of the 7s would be forced to be filled in, so i coloured those in partially lightly to see where it would take me, and that was enough to fill in the rest of the board.

Image
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Post Post #107  (ISO)  » Thu Oct 31, 2019 12:04 pm

That is correct :D
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Post Post #108  (ISO)  » Wed Jan 01, 2020 12:11 pm

I forgot that this thread existed. I made a maths geeks thread in GD (by the way, check it out!) and posted a problem there which I'm sure fits in perfectly well here. Here's the problem:

This question appeared on an exam. It went as follows:

"The product of the three integers x, y and z is 192.
z = 4, and p is equal to the average of x and y.
What is the minimum value of p?"


The four answers were: a) 6 b) 7 c) 8 d) 9,5

But... No one gave the right answer.
What is the right answer?

To make it a little bit easier in case you struggle to find the right answer, here's three hints. But don't read them unless you really need some help, and one by one!

Spoiler: Hint 1
Nobody had the right answer. Doesn't that seem a little suspicious to you?

Spoiler: Hint 2
You're probably this far that you've solved xyz=192 and z = 4 means x*y=48. Now, try to consider all possible combinations of two integers which multiply to 48.

Spoiler: Hint 3
The reason that no one gave the right answer is because there's something wrong with the question. Or the answers. Or both!
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Post Post #109  (ISO)  » Wed Jan 01, 2020 12:39 pm

Spoiler: haven't looked at your hints
7. Since z is 4, we know the product of x and y is 48. The average of x + y is minimised when they are close together. You can see this by examining the extremes: if they are 1 and 48 then the average is 24.5, if they are 6 and 8 then the average is 7.


Spoiler: after looking at hints
Ah, the answer to the question as written is -24.5. The question should have said positive integers
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Post Post #110  (ISO)  » Wed Jan 01, 2020 12:42 pm

Spoiler: Solution to vizIIsto problem
We know that xyz = 192 and z = 4, so xy = 48. In addition, p = (x + y) / 2. Since 48 is a small number, and we're dealing with integer factors, the best approach is to enumerate all the possible values of x and y snd calculate p for each case.

48 = 1 * 48 --> p = 24.5
48 = 2 * 24 --> p = 13
48 = 3 * 16 --> p = 9.5
48 = 4 * 12 --> p = 8
48 = 6 * 8 --> p = 7

Due to the commutative property of multiplication, we can ignore the remaining cases. Therefore, the answer must be b) 7. I haven't looked at the hint.

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Post Post #111  (ISO)  » Wed Jan 01, 2020 1:17 pm

You know, with the Professor Layton puzzle feeling I got by giving three hints, let's go all out by giving you guys a Layton-esque answer to your submissions.

Spoiler: Plotinus w/o hints & Ircher
WRONG.
If this was indeed the right answer, wouldn't it be striking that you would have been the only student to give the right answer?
Read the question carefully, and come to the conclusion that this answer, as obvious as may be, is incorrect.

Spoiler: Plotinus with hints
CORRECT!
No one said that the integers couldn't be negative, right?
This was later figured out by the test makers, who admitted their mistake and gave everybody the full points for the question.
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Post Post #112  (ISO)  » Mon Jan 20, 2020 8:43 pm

Maths one for you guys:
0 0 0 = 6
1 1 1 = 6
2+2+2 = 6
3 3 3 = 6
4 4 4 = 6
5 5 5 = 6
6 6 6 = 6
7 7 7 = 6
8 8 8 = 6
9 9 9 = 6

Using any mathematical symbols, make the 10 above equations true, you cannot input any extra numbers anywhere
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Post Post #113  (ISO)  » Mon Jan 20, 2020 9:12 pm

(0!+0!+0!)! = 6
(1+1+1)! = 6
2+2+2 = 6
3*3-3 = 6
sqrt(4)+sqrt(4)+sqrt(4) = 6
5+5/5 = 6
6+6-6 = 6
7-7/7 = 6
cbrt(8)+cbrt(8)+cbrt(8) = 6
sqrt(9)*sqrt(9)-sqrt(9) = 6
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Post Post #114  (ISO)  » Mon Jan 20, 2020 9:14 pm

(0!+0!+0!)! = 6
(1+1+1)! = 6
2+2+2 = 6
3!+ 3-3 = 6
(4-4/4)! = 6
5+5/5 = 6
6+6-6 = 6
7-7/7 = 6
(√(8 +8/8))! = 6
(√9)!+9-9 = 6
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Post Post #115  (ISO)  » Tue Jan 21, 2020 4:04 am

Both answers check out really well! Nice one guys
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Post Post #116  (ISO)  » Mon Feb 03, 2020 2:13 pm

Here's one I just found on YouTube:

ABC = A! + B! + C!, where ABC represents a number with A as the hundreds digit, B as the tens digit, and C as the ones digit. The digits may or may not repeat, but none of the digits are 0. Find the three-digit number ABC.
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Post Post #117  (ISO)  » Mon Feb 03, 2020 2:34 pm

Spoiler:
145.

The factorial numbers go 1, 2, 6, 24, 120, 720. The 7th is four digits long so we can cross off 720. It's 3 digits long so it has to have 120 in there, so the first digit is 1, and there's also a 5 and then it's constrained enough that you can see it at a glance
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Post Post #118  (ISO)  » Mon Feb 03, 2020 2:47 pm

Looks good :)
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Post Post #119  (ISO)  » Wed Apr 22, 2020 3:11 am

In honour of J.H. Conway.

Given any triangle ABC, extend its sides so they are lines. For each of the rays extending from each vertex, mark off a point at a distance equal to the opposite side, producing six new points, as shown below.
Image
Prove that these six points lie on a single circle.
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Post Post #120  (ISO)  » Fri May 22, 2020 8:01 pm

I think I got it...

Spoiler:
Image

Let's call the vertices of the triangle A, B, and C. And we'll call four of the extended points D, E, F, and G. I've also labeled the measures of the angles of the triangle: M_a, M_b, and M_c.

Notice that CD and EC have the same total length. This means that triangle CED is an isosceles triangle and angles CDE and DEC are equal to each other (as marked in the diagram). But we also know that the sum of the angles of triangle CED is 180 degrees. So:
180 = <DCE + <DEC + <CDE
180 = M_c + 2*<DEC
<DEC = <CDE = 90 - 0.5*M_c

Similarly, we can find that:
<BFG = <BGF = 90 - 0.5*M_b

Next, notice that triangle ADG is also isosceles, so <ADG = <AGD. Also we can see that <DAG and <BAC are equivalent angles. So <DAG = M_a and that means:
<ADG = <AGD = 90 - 0.5*M_a

Finally, calculate the sum of <FGD + <BED:
[<FGD] + [<BED]
= [(<BGF) + (<AGD)] + [<CED]
= [(90 - 0.5*M_b) + (90 - 0.5*M_a)] + [90 - 0.5*M_c]
= 270 - 0.5*M_a - 0.5*M_b - 0.5*Mc
= 270 - 0.5*(M_a + M_b + M_c)
= 270 - 0.5*180 = 270 - 90 = 180

So <FGD + <BED = 180 and it can be similarly shown that <EDG + <CFG = 180 as well. This means that quadrilateral DEFG is cyclic, so it's vertices must all fall on the same circle.

And this same method can be used to show that the other two extended points fall on the same circle as well.
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Post Post #121  (ISO)  » Fri May 22, 2020 8:06 pm

Here's a new one:

A man goes to the bank to cash a check. The cashier accidentally gives the man dollars for cents and cents for dollars by mistake. The man spends a nickel and then goes home only to notice that the amount of money that he has is double the amount of the check. How much was the check?
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Post Post #122  (ISO)  » Fri May 22, 2020 8:36 pm

Spoiler:
v is the value of the check, c is the number of cents actually received, d is the number of dollars actually received.

v = 100c + d
100d + c - 5 = 2v = 200c + 2d

98d - 199c = 5

Taking this mod 199, 98d is congruent to 5 mod 199, and 199 is prime so in the field of order 199 there's a unique solution here. 2*98 = 196 = -3, so 31 * 2 * 98 = 62 * 98 = -93. Thus, 63 * 98 = 5. So d = 63, and c = (98 * 63 - 5) / 199 = 31.

So the number of dollars actually received was 63, the number of cents was 31, and the value of the check was $31.63.

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Post Post #123  (ISO)  » Fri May 22, 2020 8:40 pm

In post 5, Ircher wrote:What exactly is a function in terms of a graph? As in, what inputs is it mapping? I may give this a try, but I would.need to know the previous.

Also, two years late, but I never answered this question on the first page!

A function on a graph typically means a function on the set of vertices. So that function maps each vertex to a nonnegative real number.

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Post Post #124  (ISO)  » Sat May 23, 2020 1:15 am

In post 120, Wickedestjr wrote:I think I got it...

Spoiler:
Image

Let's call the vertices of the triangle A, B, and C. And we'll call four of the extended points D, E, F, and G. I've also labeled the measures of the angles of the triangle: M_a, M_b, and M_c.

Notice that CD and EC have the same total length. This means that triangle CED is an isosceles triangle and angles CDE and DEC are equal to each other (as marked in the diagram). But we also know that the sum of the angles of triangle CED is 180 degrees. So:
180 = <DCE + <DEC + <CDE
180 = M_c + 2*<DEC
<DEC = <CDE = 90 - 0.5*M_c

Similarly, we can find that:
<BFG = <BGF = 90 - 0.5*M_b

Next, notice that triangle ADG is also isosceles, so <ADG = <AGD. Also we can see that <DAG and <BAC are equivalent angles. So <DAG = M_a and that means:
<ADG = <AGD = 90 - 0.5*M_a

Finally, calculate the sum of <FGD + <BED:
[<FGD] + [<BED]
= [(<BGF) + (<AGD)] + [<CED]
= [(90 - 0.5*M_b) + (90 - 0.5*M_a)] + [90 - 0.5*M_c]
= 270 - 0.5*M_a - 0.5*M_b - 0.5*Mc
= 270 - 0.5*(M_a + M_b + M_c)
= 270 - 0.5*180 = 270 - 90 = 180

So <FGD + <BED = 180 and it can be similarly shown that <EDG + <CFG = 180 as well. This means that quadrilateral DEFG is cyclic, so it's vertices must all fall on the same circle.

And this same method can be used to show that the other two extended points fall on the same circle as well.

I had a completely different proof:
Spoiler:
Let's label up.
Image

First, if the hypothesis is true, where would the centre of the circle have to be? Since UW, VY, and XZ have the same length, their closest points to the centre are equidistant. Therefore, you can draw a smaller concentric circle tangent to all three chords, and the centre must be the triangle's incentre.

With that, we go back to the triangle ABC. We find the incentre and we draw a circle so that the sides extended to chords have length equal to the perimeter of the triangle. Labeled up, we get this:
Image
Let BC = a, CA = b, and AB = c

Since the (in)centre lies on the angle bisectors of <CAB, <ABC, and <BCA, then by a symmetry argument:

AY = AZ = a'
BW = BX = b'
CU = CV = c'

Now all we need to prove is that a = a', b = b', and c = c'. This can easily be done using the equations below:

a + b' + c' = 2s(WU)
a' + b + c' = 2s(VY)
a' + b' + c = 2s(XZ)
a + b + c = 2s(definition of s)
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