Question

When 25∧25 is divided by 26 , the remainder is

Answer 1

Method (1):
Consider, $f(x)=x^{25}+1$.
As you can see $f(-1)=0$ which means $(x+1)$ is a factor of $f(x)$.
Hence, $f(x)=(x+1)g(x)$ for some $g(x)$ of degree 24.
Now, $25^{25}=(25^{25}+1)-1=f(25)-1=(25+1)g(25)-1=26(g(25)-1)+25$.
Hence, remainder is $25$.

Method (2):
From Binomial Theorem,
$25^{25}=(26-1)^{25}=^{25}C_{0}(26)^{25}+^{25}C_{1}(26)^{24}(-1)^{1}+.....+^{25}C_{25}(-1)^{25}$.
So the last term is "-1". Hence remainder is "26+(-1)" = 25.