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 B,C from A (Posted on 2007-09-14)
A(=abcde) is a 5-digit natural number with no leading zeros and e>a. B(=eca), 3-digit natural number is a multiple of 9 while C(=db), 2-digit natural number is a multiple of 3 but not 9.
Show that x5-y5=A has no integer solutions if sum of digits of C is not 6.

Note: A(=abcde) means Ten Thousand's digit of A:a, Thousand's digit of A:b and so on unit's digit of A:e.

 Submitted by Praneeth No Rating Solution: (Hide) Possible values a+c+e can take = 9,18 Possible values b+d can take = 3,12,15 So, Possible values that a+c+e-b-d can take=6,-3,-6,15,6,3. Possible values of {(a+c+e-b-d) mod 11} = 3,4,5,6,8. A=10^4 *a+10^3 *b+10^2 *c+10d+e => A mod 11 = (a-b+c-d+e) mod 11 => x^5-y^5 mod 11 = {3,4,5,6,8} ---- (1) Consider x^10 mod 11 If x is relatively prime to 11, then by Fermat's Theorem x^10 mod 11 =1 => x^5 mod 11 = +/-1. If x is a multiple of 11, then x^5 mod 11 = 0. Possible values of x^5 mod 11 ={0,+1,-1}. Possible values of x^5-y^5 mod 11={0,+1,-1,-2,+2}={0,1,2,9,10} ----(2) There are no common elements for sets 1 and 2. So, there are no solutions for x^5-y^5 = A.

Comments: ( You must be logged in to post comments.)
 Subject Author Date re:===> ergo . . . hoodat 2007-09-21 15:07:30 re(3):===>so?? [My bad] MAYBE Ady TZIDON 2007-09-20 18:36:06 re(2):===>so?? [My bad] hoodat 2007-09-20 15:34:10 re: Solution (using Excel) ===>so?? Ady TZIDON 2007-09-20 13:01:37 Solution (using Excel) hoodat 2007-09-18 13:59:46 Hint Praneeth 2007-09-17 00:32:58

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