# How do you find the remainder of 7^2002 divided by 101?

#### Answer:

The remainder is $49$

#### Explanation:

From Fermat Little Theorem we have that

${7}^{100} = 1 \left(\mod 101\right)$

but ${7}^{2002} = {7}^{2} \cdot {\left({7}^{100}\right)}^{20} = {7}^{2} \cdot 1 \mod 101 = 49 \mod 101$

So the remainder is $49$