The remainder, when $7^{103}$ is divided by 17, is __________
Answer (integer)
12
Solution
$7^{103}=7 \times 7^{102}$
<br/><br/>$$
\begin{aligned}
& =7 \times(49)^{51} \\\\
& =7 \times(51-2)^{51}
\end{aligned}
$$
<br/><br/>Remainder = $7 \times(-2)^{51}$
<br/><br/>$$
\begin{aligned}
& =-7\left(2^3 \cdot(16)^{12}\right) \\\\
& =-56(17-1)^{12}
\end{aligned}
$$
<br/><br/>Remainder $=-56 \times(-1)^{12}=-56+68=12$
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
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