find the remainder of 14^85 mod 49? i know that this is 14^85 congruent to a mod 49, where we want to find a. since (14,49) =7, it means that a is divisible by 7. then what happens next?
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note that 14 = 2*7 and 14^{85} = (2^{85})(7^{85}) = (2^{85})(7)(7^{84}) = (2^{85})(7)(7^{2})^{42} = (2^{85})(7)(49)^{42}. it stands to reason therefore that 49 divides 14^{85}, which is therefore 0 (mod 49).
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