The remainder, when $19^{200}+23^{200}$ is divided by 49 , is ___________.
Answer (integer)
29
Solution
$19^{200}+23^{200}$
<br/><br/>= $(21-2)^{200}+(21+2)^{200}=49 \lambda+2^{201}$
<br/><br/>Now, $2^{201}=8^{67}=(7+1)^{67}=49 \lambda+7 \times 67+1$
<br/><br/>$=49 \lambda+470$
<br/><br/>$=49(\lambda+9)+29$
<br/><br/>$\therefore$ Remainder $=29$
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
This question is part of PrepWiser's free JEE Main question bank. 193 more solved questions on Binomial Theorem are available — start with the harder ones if your accuracy is >70%.