Authors: Victor Sorokine
IN THE FIRST CASE every number (A) is replaced by the sum (A'+A°n) of the last digit and the remainder. After binomial expansion of the Fermat's equality, all the members are combined in two terms: E=A'^n+B'^n-C'^n with the third digit E''', which in one of the n-1 equivalent Fermat's equalities is equal to 2, and the remainder D with the third digit D''', which is equal either to 0, or to n-1, and therefore the third digit of the number A^n+B^n-C^n is different from 0.
IN THE SECOND CASE (for example A=A°n^k, but (BС)'≠0), after having transformed the 3kn-digit ending of the number B into 1 and having left only the last siginificant digits of the numbers A, В, С, simple calculations show that the (3kn-2)-th digit of the number A^n+B^n-C^n is not 0 and does not change after the restoration of all other digits in the numbers A, B, C, because it depends only on the last digit of the number A°.
Comments: 4 Pages. English version
[v1] 2019-08-05 02:01:30
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