Question

The remainder when $({15}^{23}+{23}^{23})$ is divided by 19, is:

• A. 4
• B. 15
• C. 18

Answer 1

$a^n+b^n$ is always divisible by $(a+b)$ when n is odd.

$5^{23}+{23}^{23}$ is always divisible by $15+23=38.$ As 38 is a multiple of $19,{15}^{23}+{23}^{23}$ is divisible by $19.$

Hence we get a remainder of 0.