Question

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

Solution

## We can represent $${{7}^{98}}$$ as,  $${{7}^{98}}={{\left( {{7}^{2}} \right)}^{49}}$$  $${{7}^{98}}={{\left( 49 \right)}^{49}}$$  $${{7}^{98}}={{\left( 50-1 \right)}^{49}}$$   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, $${{7}^{98}}=5k-1+5-5$$  $${{7}^{98}}=5\left( k-1 \right)+4$$   Now, divide both the sides by $$5$$. $$\dfrac{{{7}^{98}}}{5}=\dfrac{5\left( k-1 \right)}{5}+\dfrac{4}{5}$$  $$\dfrac{{{7}^{98}}}{5}=\left( k-1 \right)+\dfrac{4}{5}$$   Hence, the remainder will be $$4$$.Maths

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