Math and Logic Puzzles: Redux

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Sirius9121
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Post Post #150  (ISO)  » Sun Nov 01, 2020 3:24 am

In post 138, word321 wrote:6. What is the last digit of 7^49?



It should be 7.

Proof:
=mod(7^1,10) = 7
=mod(7^2,10) = 9
3
1
repeat (7,9,3,1)
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Post Post #151  (ISO)  » Sun Nov 01, 2020 3:27 am

4. Prove that a number is a multiple of 2 if the last digit is a multiple of 2, and that a number is multiple of 5 if the last digit is 0 or 5.


Let (number) = 10x+y where y is smallest possible interger value when x is an integer or 0.

10x = 2 * 5 * x
If y = 2m, 10x+y = 2(5x+m)

If y = 5m, 10x+y = 5(2x+m)
If you ask Rick Astley for his copy of the movie 'Up', he can not give it to you as he will never give you up.
However, by refusing to do so, he lets you down.
Thus creating the Astley Paradox.
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Post Post #152  (ISO)  » Sun Nov 01, 2020 3:31 am

Prove that a number is multiple of 3 if the sum of its digits is a multiple of 3


Let x = a * 10^1 + b * 10^2 + c * 10^3 ... ? * 10^(digits of x)

x = a * (10^1-1) + b * (10^2-1) ... ? * 10^(digitsofx) + (a + b + c...)
Since (10^integer-1) must be 9 repeated (integer) times, (10^integer-1) must be a multiple of 3

x = 3(something) + (a + b + c...)

if a + b + c... = 3n, x = 3(something+n)
If you ask Rick Astley for his copy of the movie 'Up', he can not give it to you as he will never give you up.
However, by refusing to do so, he lets you down.
Thus creating the Astley Paradox.
*glares at Sirius*Well played my friend, well played. - Aristophanes

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Post Post #153  (ISO)  » Tue Dec 08, 2020 11:48 pm

Let us assume we roll a fair dice a very large number of times. What is the probability that more (123s) than (456s) are rolled and ALSO have more (135s) than (246s)?
Last edited by Sirius9121 on Wed Jan 06, 2021 8:08 am, edited 1 time in total.
If you ask Rick Astley for his copy of the movie 'Up', he can not give it to you as he will never give you up.
However, by refusing to do so, he lets you down.
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Post Post #154  (ISO)  » Wed Dec 09, 2020 4:49 am

I suspect it's 1/2 but am not confident and am intrigued to try to show it formally and will think about it more later

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Post Post #155  (ISO)  » Wed Dec 09, 2020 4:56 am

In post 154, implosion wrote:I suspect it's 1/2 but am not confident and am intrigued to try to show it formally and will think about it more later

please note that both requirements need to be fulfilled... its lower than 1/2
If you ask Rick Astley for his copy of the movie 'Up', he can not give it to you as he will never give you up.
However, by refusing to do so, he lets you down.
Thus creating the Astley Paradox.
*glares at Sirius*Well played my friend, well played. - Aristophanes

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Post Post #156  (ISO)  » Wed Dec 09, 2020 8:27 am

Still waking up, but I believe it's 1/4. Will write out a full proof later if someone hasn't by then.
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Post Post #157  (ISO)  » Wed Dec 09, 2020 8:31 am

I feel like it’s going to be in the ballpark of 1/3. The two things we need aren’t quite independent (2/3 of the small numbers are odd whereas 1/3 of the big numbers are odd).
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Post Post #158  (ISO)  » Wed Dec 09, 2020 10:24 am

Oh, yeah, missed that.
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Post Post #159  (ISO)  » Wed Dec 09, 2020 2:18 pm

In post 155, Sirius9121 wrote:
In post 154, implosion wrote:I suspect it's 1/2 but am not confident and am intrigued to try to show it formally and will think about it more later

please note that both requirements need to be fulfilled... its lower than 1/2

I understand, I came up with a heuristic argument that made me think it would be 1/2 anyway asymptotically. But it certainly wasn't rigorous or anything.

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Post Post #160  (ISO)  » Wed Dec 09, 2020 4:25 pm

Making a few assumptions about the relevant probability distribution, I get P = arctan(sqrt(2))/pi ~= 30.4%. Those assumptions could very easily be wrong though.
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Post Post #161  (ISO)  » Wed Dec 09, 2020 6:22 pm

I get the same result, also making some assumptions, but I feel reasonably confident in them.

How I get it:
Let A be the number of times a 1 is rolled minus the number of times a 4 is rolled, let B be the number of times a 3 is rolled minus the number of times a 6 is rolled, let C be the number of times a 2 is rolled minus the number of times a 5 is rolled. The event is now:
A+B>|C|
Also note that A, B, and C are identically distributed and approximately normal with mean 0 via CLT. Normalizing to make them variance 1 doesn't change the equation, so if we assume they are independent it's now
X>|Y| where X is normal with mean 0 variance 2, Y is normal with mean 0 variance 1. We then do basic integration (Set up the integral, note that it's the standard e^(-r^2/2) with r from 0 to infinity but over a sector with a weird angle) and end up with the answer scigatt gave. A,B, and C are obviously not independent, for example if A=N then that means that every roll was a 1 so B and C must both be 0. But they are [assumption] approximately independent because that is unlikely to happen.
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Post Post #162  (ISO)  » Sat Dec 12, 2020 6:38 pm

I believe that a more rigorous approximation for P is the following:

Image

I think from this the limit value of arctan(sqrt(2))/pi follows easily.
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Post Post #163  (ISO)  » Wed Dec 23, 2020 9:28 pm

How I've been keeping busy during a pandemic...






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Post Post #164  (ISO)  » Mon Dec 28, 2020 2:15 pm

Here's one I saw somewhere: given that x,y are chosen randomly uniformly in [0,1], what is the probability that floor(x/y) is even?
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Post Post #165  (ISO)  » Mon Dec 28, 2020 5:54 pm

Spoiler: "Solution for e trillion's problem"
Let's see if this works: Take the unit square [0, 1] X [0, 1], which has area 1. The points corresponding to floor(x/y) being even within this square are as follows:
- Everything to the left (or above) of y = x.
- Everything between x = 2y and x = 3y.
- Everything between x = 4y and x = 5y.
etc., forming triangles between the lines x = 2ky and x = (2k + 1)y, for positive integers k.
These triangles will have an apex at (0, 0) and the corresponding base on the line x = 1. The end-points of the base will be at 1/(2k) and 1/(2k + 1) respectively. In other words, the area of the kth triangle will be (1/2) * [1/(2k) - 1/(2k + 1)]. Note that the first triangle (left of y = x) is not included in this calculation and has area 1/2.
Therefore, the total area of the triangles will be 1/2 + Sum (k = 1 to infinity) (1/2) * [1/(2k) - 1/(2k + 1)] = (1/2) * [2 + Sum (n = 1 to infinity) [(-1)^n]/n]. This sum is the negative of the alternating harmonic series, so the area comes to (1/2) * (2 - ln 2) = 1 - (ln 2)/2, which is the probability requested and approximately 65.34%.
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Post Post #166  (ISO)  » Tue Dec 29, 2020 2:59 am

In post 163, mith wrote:How I've been keeping busy during a pandemic...

great channel, got introduced earlier this year as a way to destress. it's very relaxing to watch the solves. i can only do the easy puzzles though, if it's a little hard i always get completely stumped.
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Post Post #167  (ISO)  » Tue Dec 29, 2020 5:35 am

Mitillos's solution seems to be correct.
Spoiler: why
Image
This is me, running this simulation 500000 times by randomizing x and y then dividing them. The probablity by this method was around 65.3574%. Since this is a completely random simulation, it makes sense to have the result differ a bit, therefore I agree with Mitillos's solution.
If you ask Rick Astley for his copy of the movie 'Up', he can not give it to you as he will never give you up.
However, by refusing to do so, he lets you down.
Thus creating the Astley Paradox.
*glares at Sirius*Well played my friend, well played. - Aristophanes

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