4 ms·
Fermat Theorem Puzzle
- datr 14y agoSolution: Look at the equation modulo 10.
- scovetta 14y agoHere's my solution: x=2233445566 y=7788990011 z=9988776655 6 * 6 = 36 (last digit stays the same) 1 * 1 = 1 (last digit stays the same) 5 * 5 = 25 (last digit stays the same) so: x^N ends with 6 y^N ends with 1 z^N ends with 5 for just that last digit, 6 + 1 != 5. qed
- HoLyVieR 14y agoHere's an other solution. If you simply calculate the first digit and length for N=2, you will see that 2233445566 has a length of 19 and starts with 4, 7788990011 has a length of 20 and starts with 4 and 9988776655 has a length of 20 and starts with 8. If the first number had a length of 20 it could match, but since it has a length of 19 no mater the rest of the digits it can't be equal.
- twiceaday 14y agoLooking at the equation mod 3 you get 1 + 2^N = 1 mod 3 or 2^N = 0 mod 3. Not possible.
- tzs 14y agoIt's easy to show, using the binomial theorem, that if x < y and x^n + y^n = z^n, with x, y, z, n positive integers, then we must have n(z-y) <= x. For the given numbers, this puts an upper limit on n of 1, and so x, y, z cannot be a Fermat counterexample.
- keymone 14y agoit's a puzzle for 10yr olds