The remainder, when $7^{103}$ is divided by 23, is equal to:
Solution
<p>$$\begin{aligned}
& 7^{103}=7\left(7^{102}\right)=7(343)^{34}=7(345-2)^{34} \\
& 7^{103}=23 \mathrm{~K}_1+7.2^{34} \\
& \text { Now } 7.2^{34}=7 \cdot 2^2 \cdot 2^{32} \\
& =28 \cdot(256)^4 \\
& =28(253+3)^4 \\
& \therefore 28 \times 81 \Rightarrow(23+5)(69+12) \\
& 23 \mathrm{~K}_2+60 \\
& \therefore \text { Remainder }=14
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
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