By the remainder theorem (also known as the factor theorem), non-zero (a - b) divides (a^n - b^n). For odd n, we therefore have (a - (-b)) divides (a^n - (-b)^n), or (a + b) divides (a^n + b^n).
For odd n, we can also explicitly factorize a^n + b^n, as follows. Writing n = 2m+1, we have:
a^(2m+1) + b^(2m+1) = (a + b)(a^2m - a^(2m-1)b + a^(2m-2)b^2 - ... - ab^(2m-1) + b^2m).
Rewriting the series as (1^99 + 99^99) + (2^99 + 98^99) + ... + (49^99 + 51^99) + (50^99), 100 divides each of the first 49 bracketed terms by the factor theorem. Since 100 divides 50^2, it also divides 50^99.
Therefore 100 divides the whole series, and the last two digits are 00.
Edited on January 31, 2004, 7:53 am
Edited on January 31, 2004, 8:06 am
Last two digits
