
Fermat Theorem Puzzle - pratikpoddar
http://pratikpoddarcse.blogspot.in/2013/01/fermat-theorem-puzzle.html
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HoLyVieR
Here'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.

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scovetta
Here'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

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tzs
It'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.

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twiceaday
Looking at the equation mod 3 you get 1 + 2^N = 1 mod 3 or 2^N = 0 mod 3. Not
possible.

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datr
Solution: Look at the equation modulo 10.

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koopajah
You can also use the last digit of each number. x^N will always end with a
'6', y^N with a '1' and z^N with a '5' for N>=1 so ut can't work

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twiceaday
This is equivalent to looking at the equation modulo 10.

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keymone
it's a puzzle for 10yr olds

