The remainder when $(11)^{1011}+(1011)^{11}$ is divided by 9 is
Solution
<p>$${\mathop{\rm Re}\nolimits} \left( {{{{{(11)}^{1011}} + {{(1011)}^{11}}} \over 9}} \right) = {\mathop{\rm Re}\nolimits} \left( {{{{2^{1011}} + {3^{11}}} \over 9}} \right)$$</p>
<p>For ${\mathop{\rm Re}\nolimits} \left( {{{{2^{1011}}} \over 9}} \right)$</p>
<p>$${2^{1011}} = {(9 - 1)^{337}} = {}^{337}{C_0}{9^{337}}{( - 1)^0} + {}^{337}{C_1}{9^{336}}{( - 1)^1} + {}^{337}{C_2}{9^{335}}{( - 1)^2} + \,\,.....\,\, + \,\,{}^{337}{C_{337}}{9^0}{( - 1)^{337}}$$</p>
<p>So, remainder is 8 </p>
<p>and ${\mathop{\rm Re}\nolimits} \left( {{{{3^{11}}} \over 9}} \right) = 0$</p>
<p>So, remainder is 8</p>
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
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