Question

Find the reminder, when $7^{98}$ is divided by $5$

Answer 1

We can represent $7^{98}$ as,

$7^{98}={\left(7^2\right)}^{49}$

$7^{98}={\left(49\right)}^{49}$

$7^{98}={(50-1)}^{49}$

Therefore, using binomial expansion, we have

$7^{98}{=}^{49}{C}_0{\left(50\right)}^{49}{-}^{49}{C}_1{\left(50\right)}^{48}+.......{+}^{49}{C}_{48}\left(50\right){-}^{49}{C}_{49}$

Here, except the last each term contains multiple of $5$. Therefore,

$7^{98}=5\left(k\right)-1$, where $k$ be any integer

Now,

$7^{98}=5k-1+5-5$

$7^{98}=5(k-1)+4$

Now, divide both the sides by $5$.

$\frac{7^{98}}{5}=\frac{5(k-1)}{5}+\frac{4}{5}$

$\frac{7^{98}}{5}=(k-1)+\frac{4}{5}$

Hence, the remainder will be $4$.