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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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