Fermat Theorem Puzzle


Source: Andrej Cherkaev's List of Puzzles



Problem:
A computer scientist claims that he proved somehow that the Fermat theorem is correct for the following 3 numbers:

x=2233445566,
y=7788990011,
z=9988776655

He announces these 3 numbers and calls for a press conference where he is going to present the value of N (to show that

x^N + y^N = z^N

and that the guy from Princeton was wrong). As the press conference starts, a 10-years old boy raises his hand and says that the respectable scientist has made a mistake and the Fermat theorem cannot hold for those 3 numbers. The scientist checks his computer calculations and finds a bug.

How did the boy figure out that the scientist was wrong?

Update (06/01/2012):
Solution posted by a lot of people in comments!

Comments

  1. By last digit number? Or is it even easier than that?

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  2. Any power of x = 2233445566 must end with the digit 6 and any power of y = 7788990011 must end with the digit 1. Further, any power of z must also end with 5. Hence, x^N + y^N can not be equal to z^N for any value of N.

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  3. last digit of x^n + y^ n = 7, and last digit of z^n = 5.

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  4. x^n for any value of n will always have the least significant digit as 6, similarly y^n for any value of n will always end with 1 , and hence there sum will always end with the least significant digit as 7 , whereas z^n will always have 5 as the least significant digit.

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  5. whatever the value of N ,x^N will always end with 6(one's place) and. y^N will end with 1. so there sum can never give number with a one's digit equal to 5 (which will always be the case for z^N)

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  6. 11 can be factored out from x,y,z.
    we get X = 203040506 = 203040500 + 6
    Y = 708090001 = 708090000 + 1
    Z = 908070605 = 908070600 + 5

    The way X, Y, Z have been presented it is clear that X raised to N + Y raised to N will have 7 in its units place. ,
    where as Z will have 5 ,
    they can never be equal

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  7. Claim:
    xxxxx6^N = 6 mod 10 for all N
    xxxxx1^N = 1 mod 10 for all N
    xxxxx5^N = 5 mod 10 for all N

    Proof:
    By induction
    Let us suppose xxxx6^(N-1) = 6 mod 10
    xxxxx6 = 6 mod 10
    Thus xxxx6^N = 6*6 mod 10 = 6 mod 10

    Similarly, the claim can be proved for the other two cases.

    if xxxx6^N + xxxxx1^N = xxxxx5^N
    then xxxx6^N % 10 + xxxxx1^N % 10 = xxxxx5^N % 10 which is a contradiction from the above claim

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  8. Any power of x ends with a 6, that of y ends with a 1 and that of z ends with 5 . Thus x^n +y^n ends with 7 while z^n ends with 5. This proves that the scientist was wrong

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  9. The units digit of x^N is 6, of y^N is 1, and of z^N is 5. [*]

    The units digit of x^N + y^N is therefore 7. Given what we know about the units digit of z^N, the scientist must therefore be wrong.

    [*] Consider (10a+6)² = (100a² + 20a + 36) and apply in an inductive proof. Same method for (10a+5) and (10a+1).

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  10. x^anything will end with 6, y^anything will end with 1. So their sum should end with 7. z^anything should end with 5.

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  11. This one's seems easy.

    Since units digit of x is 6, the units digit of x^n will be 6 irrespective of the value of n.
    Similarly units place of y^n will always be 1 and that of z^n will always be 5.

    So the units digit of LHS is 7 and that of RHS is 5. Clear mismatch.

    Hence that boy could prove the great scientist wrong using just simple school maths !!

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  12. we can see that the equality can't hold by observing the equality mod 10. x^N % 10 = 6 y^N%10 = 1 and z^N%10 = 5 always.

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  13. Um, you mean that the scientist claims to have a *counterexample* to Fermat's Last Theorem. I'll keep quiet about the solution.

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  14. Does not even work for N=1

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  15. Can you please provide the source please?

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  16. This comment has been removed by the author.

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  17. 6^n always ends in a 6.. 1^n always ends in a 1 .. and 5^n always ends in a 5..

    With the given numbers .. x^n+y^n must end in 7 ..

    Hence, there is surely some mistake ..

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  18. x ^ N mod 10 = 1
    y ^ N mod 10 = 6

    and

    z ^ N mod 10 = 5

    So equation will never hold modulo 10

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  19. shouldn't x^N+y^N = 7 mod 10 ?

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  20. hint: any power of a number ending in 6 will always end in 6.

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  21. With following argument :
    -> every exponent of x has unit digit 6
    -> every exponent of y has unit digit 1

    so (x^N + y^N) will have unit digit 7

    -> but every exponent of z has unit digit 5

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  22. @Ameya, @Anirudh Kumar, @Siddhartha, @donotevenbother, @iron feliks, @vivek ranjan nema, @napster, @aditya_iitr, @JDGM, @Vivek, @amit yadav, @rishabh, @r, @akhil, @piyush, @bob, @shubham.. Thanks for the correct solution :-)

    @andrew milne..
    x+y>z, but x^n + y^n can be equal to z^n. The scientist has to come up with the number n. To disprove Fermat's last theorem, you had to provide that there exists a 'N'

    @ravi. source is already provided

    @as. The scientist has to come up with the number n. To disprove Fermat's last theorem, you had to provide that there exists a 'N'


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  23. Based on the example x+y > z. So there doesn't exist any N>2. ( P.S. I get the essence of question )

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  24. Here is a different solution:
    x is 1 mod 3, y is -1 mod 3. So x^n + y^n will always be either 0 mod 3 or 2 mod 3.
    However, z is 1 mod 3, and hence z^n is always 1 mod 3.

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    Replies
    1. clearly, it doesn't work for N=1. Now for N>1 and given X and Y, we have
      X^N > X ; Y^N>Y => X^N+Y^N = Z > X+Y;
      thus Z > 10022435577 for all N >1 ; The scientist provides Z as 9988776655 which clearly violates this. Thus by simple 10 yr old logic, the scientist was wronged.

      Delete
    2. Kunal, the equation is X^N+Y^N = Z^N and not X^N+Y^N = Z

      Delete

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