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> Find the remainder using Fermat's little theorem when $5^{119}$ is divided by $59$.
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Fermat's little theorem states that if $p$ is prime and $\operatorname{gcd}(a,p)=1$,then $a^{p-1} -1$ is a multiple of $p$.

For example, $p=5,a=3$. From the theorem, $3^5-1 -1$ is a multiple of $5$ i.e $80$ is a multiple of $5$.

Similarly, I need to find the remainder when $5^{119}$ is divided by $59$.

**My approach:**

Using theorem I solve:

$5^{119-1} -1=5^{118} - 1 \Rightarrow 5^{118}=59k+1$, where $k$ is a natural number.

How do I proceed?