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Math Puzzles/Problems

AndrewAndrew
edited September 2008 in Forum Game
Rules of the thread:
  • This thread for posting and solving math related puzzles and problems. All forms of math welcome.
  • Show your work ~_^
  • If you solve a problem, you must post a new one.
  • More than one problem can be open/unsolved at a time. Try to update your posts to reflect if a problem has been solved.
  • Try to stay on topic (as best as could be expect in this forum).
  • Don't be afraid to ask for help!
  • Contribute the best you can, easy and hard problems are both welcome.
  • Sudoku is not math, so get over yourself.
  • Don't post a problem that you don't know the solution to. (No, we will not do your homework.)
First Problem:
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
«13456

Comments

  • EsonLinjiEsonLinji
    I might be missing something but isn't it just two? {1,25} would be the first of twelve such solutions. Is there any other constraints on choosing the integers?

    For those who want lots of problems, there's a bunch of maths/computer programming challenges over at http://projecteuler.net
  • StarfoxStarfox
    edited September 2008
    I don't understand the question.

    Yeah, what he said. ^
    Post edited by Starfox on
  • AndrewAndrew
    I might be missing something but isn't it just two? {1,25} would be the first of twelve such solutions. Is there any other constraints on choosing the integers?
    I'm looking for the number of integers that must be selected from the set to guarantee that two of them add up to 26. I.E If I pick two integers, they may not add up to 26 (could be 1 and 25 OR they could be 2 and 3). How many numbers must I pick at minimum to guarantee that two of them add up to the correct sum?
  • StarfoxStarfox
    I believe it's 13. If you select {25, 24, 23, ... 14,13}, you have a set of size 12, and the smallest it can add up to is 27. If you pick one more number anywhere lower, it must add up to 26 with one of the first 12 you picked.
  • TechparadoxTechparadox
    edited September 2008
    I'm going to preface this by saying I suck at math problems involving probability. I'm guessing the number of integers you'd have to pick from a set of 1-25 to insure a guarantee that two of them would add up to 26 would be 14. I'm basing this on the theory that you could hit 13 random numbers that don't match up for squat, but as soon as you get to the 14th one it's going to have to match up to one of the other 13 and trip the ending condition of two of the numbers adding up to 26. Or to put it another way, (.5 x S) + 1, rounded to the nearest whole number. You could run through numbers 1-13 without having the finishing condition, but as soon as you hit 14 the condition is met (14+12=26).
    Post edited by Techparadox on
  • AndrewAndrew
    edited September 2008
    I believe it's 13. If you select {25, 24, 23, ... 14,13}, you have a set of size 12, and the smallest it can add up to is 27. If you pick one more number anywhere lower, it must add up to 26 with one of the first 12 you picked.
    Very close, but you made a counting error.
    I'm guessing the number of integers you'd have to pick from a set of 1-25 to insure a guarantee that two of them would add up to 26 would be 14.
    Correct! ^_^
    Post edited by Andrew on
  • StarfoxStarfox
    edited September 2008
    Drat.

    I didn't have a problem to pose anyway. Wait, here's one:
    P = NP? Unsolved.
    Post edited by Starfox on
  • JasonJason
    Jason has a dozen donuts. He eats two of them and then Andrew steals one.

    How hard does Jason kick Andrew's ass? Answer must be in pounds per square inch.
  • AprecheApreche
    edited September 2008
    Jason has a dozen donuts. He eats two of them and then Andrew steals one. How hard does Jason kick Andrew's ass? Answer must be in pounds per square inch.
    imageppsi
    It doesn't get harder than that.
    Post edited by Apreche on
  • AndrewAndrew
    How hard does Jason kick Andrew's ass? Answer must be in pounds per square inch.
    Nothing. Because Jason eats a shit ton of donuts and is a man of larger stature, he waddles after Andrew to no avail.
  • SailSail
    P = NP?
    N=1? That seems too obvious.
  • StarfoxStarfox
    Somebody doesn't get the problem...
  • JasonJason
    Ug-ugugugugug.
    image
  • SailSail
    edited September 2008
    Somebody doesn't get the problem...
    Clearly.

    [Edit] Oh. Thank you Wikipedia.
    Post edited by Sail on
  • TechparadoxTechparadox
    edited September 2008
    I'm still trying to come up with a decent one - bear with me.
    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)
    Sudoku is not math, so get over yourself.
    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.
    Post edited by Techparadox on
  • ΣπεκωσποκΣπεκωσποκ
    I'm still trying to come up with a decent one - bear with me.
    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?

    Unsolved(and I'll be honest, I didn't come up with this one myself)

    Sudoku is not math, so get over yourself.
    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.He has a 25/44 chance in losing. For the first five rounds, he and his opponent have 19 bullets total to avoid, but the first player is guaranteed to lose at the start of the sixth round. Thus, one adds 6 bullets to the first's total, and 0 to the second's total, making the total number of bullets to be avoided 25 for the first, and 19 for the second, for a total of 44 bullets to be avoided.
  • Luke BurrageLuke Burrage
    I'm guessing the number of integers you'd have to pick from a set of 1-25 to insure a guarantee that two of them would add up to 26 would be 14.
    Correct! ^_^
    This is just a variation of the "I have x number of red socks in a drawer and y number of blue socks, how many socks do I need to take out of the drawer to be holding a pair?" Right?
  • BronzdragonBronzdragon
    He has a 25/44 chance in losing. For the first five rounds, he and his opponent have 19 bullets total to avoid, but the first player is guaranteed to lose at the start of the sixth round. Thus, one adds 6 bullets to the first's total, and 0 to the second's total, making the total number of bullets to be avoided 25 for the first, and 19 for the second, for a total of 44 bullets to be avoided.
    Rules of the thread:
    • If you solve a problem, you must post a new one.
  • TechparadoxTechparadox
    edited September 2008
    He has a 25/44 chance in losing. For the first five rounds, he and his opponent have 19 bullets total to avoid, but the first player is guaranteed to lose at the start of the sixth round. Thus, one adds 6 bullets to the first's total, and 0 to the second's total, making the total number of bullets to be avoided 25 for the first, and 19 for the second, for a total of 44 bullets to be avoided.
    I can see your logic there, but that doesn't match the solution I got from the webpage where I found the problem. It's darned close, though.

    Also, in a Russian-Roulette-related vein, check this out.
    Post edited by Techparadox on
  • lackofcheeselackofcheese
    edited September 2008
    He has a 25/44 chance in losing. For the first five rounds, he and his opponent have 19 bullets total to avoid, but the first player is guaranteed to lose at the start of the sixth round. Thus, one adds 6 bullets to the first's total, and 0 to the second's total, making the total number of bullets to be avoided 25 for the first, and 19 for the second, for a total of 44 bullets to be avoided.
    Rules of the thread:
    • If you solve a problem, you must post a new one.
    It doesn't say you must post a new one instantly, nor in the same post as the one in which you have solved the problem.
    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, then
    There 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.
    Post edited by lackofcheese on
  • TechparadoxTechparadox
    edited September 2008
    *DING* We have a winner!

    Again, I suggest checking out that link I posted. Princess Bride fans will enjoy it, too.
    Post edited by Techparadox on
  • lackofcheeselackofcheese
    I couldn't help answering, but now I have to come up with a new problem :S
  • Dr. TimoDr. Timo
    edited September 2008
    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?
    Assuming you don't spin the chamber between rounds:
    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%.
    Post edited by Dr. Timo on
  • Dr. TimoDr. Timo
    edited September 2008
    Dang took me too long to write up my answer :). Oh well. I do have some nice problems though.
    Post edited by Dr. Timo on
  • lackofcheeselackofcheese
    Heh; go ahead with yours, it's better than waiting for me to come up with one.
  • Dr. TimoDr. Timo
    edited September 2008
    OK, here is one of my favorites:
    --
    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 ;)
    Post edited by Dr. Timo on
  • Luke BurrageLuke Burrage
    edited September 2008
    Do the prisoners know when a day is up?

    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.
    Post edited by Luke Burrage on
  • ......
    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.
    I think that's the correct answer yes. As for the prisoners knowing when a day is up, their cells are windowless and soundproof, so I doubt they can, and it wouldn't help them though, the living room visiting prisoner is random, so its possible that one prisoner has spend 3 days in the living room, and another hasn't spend a single day there.
  • HungryJoeHungryJoe
    For integers n such that n > 2, does the equation an + bn = cn 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.
  • StarfoxStarfox
    Now Joe, you're breaking the same rule I did. ^_~
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