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- This thread for posting and solving math related puzzles and problems. All forms of math welcome.
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How many integers from the set S={1,2,3,4,5,...,23,24,25} must be chosen so at least two of them add up to a sum of 26? Solved by Techparadox

## Comments

For those who want lots of problems, there's a bunch of maths/computer programming challenges over at http://projecteuler.net

Yeah, what he said. ^

I didn't have a problem to pose anyway. Wait, here's one:

P = NP? Unsolved.

How hard does Jason kick Andrew's ass? Answer must be in pounds per square inch.

It doesn't get harder than that.

[Edit] Oh. Thank you Wikipedia.

Ok, I have one that's probability-related :

Assuming both players take turns what is the probability the player who goes first will lose at Russian roulette using a gun with six chambers?

lackofcheese (and I'll be honest, I didn't come up with this one myself) Too true. I once heard a co-worker say he didn't like Sudoku because he wasn't good at math. I looked right at him and said, "What math? All you have to do is be able to count to nine and use a little logic." He wasn't impressed or swayed by my take on it.

Also, in a Russian-Roulette-related vein, check this out.

Verification of the solution is also absent, though it doesn't seem right to me.

It depends on the specifics of the Russian Roulette involved.

If it's played such that there is always one bullet in the chamber, and the chamber is spun each time, then:The probability of your losing on the first shot is 1/6.

If you survive a shot (probability 5/6), the probability of your losing on the next shot is (5/6 (enemy doesn't lose) * 1/6)

Hence the probability of your losing overall is

1/6 + (5/6 * 5/6 * 1/6) + (5/6 * 5/6 * 5/6 * 5/6 * 1/6) + ...

= 1/6 * (1 + (25/36) + (25/36)^2 + ...)

This is an infinite geometric series, the value of which is

a / (1 - r)

=(1/6) / (1 - 25/36)

=(1/6) / (11/36)

= (1/(11/6))

= 6/11

If it's played such that there is always one bullet in the chamber, and the chamber is not spun, thenThere are 6 locations the bullet could have been in initially. 3 mean the first person dies, the other three mean the second person dies.

Hence the probability is 1/2.

Again, I suggest checking out that link I posted. Princess Bride fans will enjoy it, too.

There are six initial states the gun can be in. These states have equal probability. For half of these states the first player will lose. The answer is 50%.

Assuming you spin the chamber in between rounds:

First time lethality: 1/6

Second time lethality: 5/6 (it didn't kill the first player) x 5/6 (it didn't kill the second player) x 1/6 (it kills the first player the second time it comes to him)

n:th time lethality: 1/6(Sum[(5/6)^(2m), {m, 0, n}]) -> 0.54(54 repeat) as n --> infinity.

The answer is roughly 55%.

--

100 prisoners are imprisoned in solitary cells. Each cell is windowless and soundproof. There's a central living room with one light bulb; the bulb is initially off. No prisoner can see the light bulb from his or her own cell. Each day, the warden picks a prisoner equally at random, and that prisoner visits the central living room; at the end of the day the prisoner is returned to his cell. While in the living room, the prisoner can toggle the bulb if he or she wishes. Also, the prisoner has the option of asserting the claim that all 100 prisoners have been to the living room. If this assertion is false (that is, some prisoners still haven't been to the living room), all 100 prisoners will be shot for their stupidity. However, if it is indeed true, all prisoners are set free and inducted into MENSA, since the world can always use more smart people. Thus, the assertion should only be made if the prisoner is 100% certain of its validity.

Before this whole procedure begins, the prisoners are allowed to get together in the courtyard to discuss a plan. What is the optimal plan they can agree on, so that eventually, someone will make a correct assertion?

--

So, this has an easy solution and a hard one. I'll accept the easy solution and applaud the hard one.

Edit: solved by Luke (the easy case). The solution will take O[100^2] ~= 10000 days ~= 30 years. The expected time for everyone to have been in the room is 100 Log[100] = 460 days, so the solution is quite wastefull. The difficult solution takes around 3500 days.

Now you can solve the same problem with two light bulbs, indefinite initial state and the option for the jailer to randomly skip days

Edit:

Ok, one person does nothing but turn the lights off. Everyone else turns the lights on once, and if they come into the room and the light is already on, that doesn't count for them. Once the first person has turned off the light 100 times he knows everyone has been out. I think this could take too long, but you didn't say the life expectancies of anyone.

Edit 2:

I don't have any math workings to back up my answer.

^{n}+ b^{n}= c^{n}have any solutions in non-zero integers a, b, and c? Prove it.Can every even integer greater than 2 can be written as the sum of two primes? If so, prove it.