JEE MAIN - Mathematics (2022 - 25th July Evening Shift - No. 4)

The remainder when $$(11)^{1011}+(1011)^{11}$$ is divided by 9 is
1
4
6
8

Explanation

$${\mathop{\rm Re}\nolimits} \left( {{{{{(11)}^{1011}} + {{(1011)}^{11}}} \over 9}} \right) = {\mathop{\rm Re}\nolimits} \left( {{{{2^{1011}} + {3^{11}}} \over 9}} \right)$$

For $${\mathop{\rm Re}\nolimits} \left( {{{{2^{1011}}} \over 9}} \right)$$

$${2^{1011}} = {(9 - 1)^{337}} = {}^{337}{C_0}{9^{337}}{( - 1)^0} + {}^{337}{C_1}{9^{336}}{( - 1)^1} + {}^{337}{C_2}{9^{335}}{( - 1)^2} + \,\,.....\,\, + \,\,{}^{337}{C_{337}}{9^0}{( - 1)^{337}}$$

So, remainder is 8

and $${\mathop{\rm Re}\nolimits} \left( {{{{3^{11}}} \over 9}} \right) = 0$$

So, remainder is 8

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