what would be the remainder when 35^12+63^17 is divided by 14? I am unable to understand how I should approach this problem.
35^{12} + 63^{17} = 7^{12} \left (5^{12} + 7^5 \cdot 9^{17}\right)
Can you finish the problem from here? As a hint, we want to consider the parity of the term in the parentheses.
in the parentheses 5 is Odd and 7 (odd) * 9 odd) is also odd. Therefore Odd+Odd is even and we can cancel out 2 from 14 in the denominator. Since we are only left with seven in the denominator and 7^12 in the numerator, remainder should be 0?
“canceling out” isn’t exactly the correct terminology, but yes you have the right idea. The “numerator” (35^{12} + 63^{17}) would be a multiple of 14, and as you know, its remainder when divided by 14 must be 0.