when 4^101+6^101 is divided by 25, the remainder is:
(A)20 (B)10 (C)5 (D)0
recall that for odd powers,
a^n + b^n = (a+b)(a^(n-1) - a^(n-2)*b + ...)
since 4+6=10, the product is a multiple of 10, so dividing by 5 leaves a remainder of zero.
(D)