More mathy brainteasers!
So if A^n+B^n=C^n
, we all know there are many cases for n=2 (right triangles...)
but for n>2, there are no solutions.

Prove it.

Anyone that knows anything about math/this problem: don't say anything. Just curious as to how people approach this problem/whether it causes funny arguments.

Are you kidding me? Who the fuck gives this at an interview? I'd punch them in the face and walk out.

After solving it of course.

Just Do It

Trivial by inspection

I think this one may be too difficult for non-maths/cs people

Best Response

I have discovered a truly marvelous proof of this, but it's too long for this comment box.

Don't believe everything you think.

joe_monkey:

I have discovered a truly marvelous proof of this, but it's too long for this comment box.

lol, well done.

true wasn't wile's proof like 200 pages or something, gl getting through that in an interview

HA ha ha ah ah......I'm sure NO ONE got asked this in an interview.

The proof is like 100 pages long, it's probably the most argued theorem in math, and way beyond even some PhD's .

+1 SB to Joe Monkey for quoting the person who first claimed to solve this interview brainteaser.

Yeah, this was a social experiment. Didn't expect this many people to be familiar with Fermat's last theorem/wiles' proof on WSO.

And yeah, you need really advanced graduate level math to even begin understanding the proof. You have to admit, with the mathematical ineptitude displayed in some of the brain teaser threads, it wasn't that farfetched to think most people here wouldn't recognize it on sight.

This logic puzzle, on the other hand, actually is very hard. It's not conceptually a brain teaser, per se, but requires really deep logical thought. You practically need to do everything based on symbolic logic.

"Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are da and ja, in some order. You do not know which word means which.
"
http://en.wikipedia.org/wiki/The_Hardest_Logic_Puz...

http://math.stanford.edu/~lekheng/flt/wiles.pdf

Have fun!

"Life all comes down to a few moments. This is one of them." - Bud Fox

You have an unlimited supply of quarters, dimes, nickels, and pennies. What is the greatest value you can reach such that the coins that you have chosen cannot be rearranged to equal exactly \$1.00?

al865149:

You have an unlimited supply of quarters, dimes, nickels, and pennies. What is the greatest value you can reach such that the coins that you have chosen cannot be rearranged to equal exactly \$1.00?

solution: imagine there is a bowl in front of you. take the bowl, your coins, and carry out the following steps:

[step 1] pick a denomination, d, that has not yet been selected. if there are no denominations left to choose from, stop. otherwise, go to step 2.

[step 2] determine the maximum number, n, of coins of denomination d that you can put in the bowl such that you cannot, even with the change in bowl from any previous steps, make change for a dollar. then, go to step 3.

[step 3] put n coins of denomination d in the bowl. then, go to step 1.

so, if i go dimes, quarters, nickels, then pennies, we get: 9 dimes + 1 quarter + 0 nickels + 4 pennies. no matter which denomination you start with, the max is \$1.19

Didn't wiles present at some conference and it took various hours to go through it all?