A) 3
B) 7
C) 9
D) 11
Correct Answer: C
Solution :
\[2{{7}^{40}}={{3}^{120}}\] \[{{3}^{119}}=\,\,{{\left( 4-1 \right)}^{119}}{{=}^{119}}{{C}_{0}}{{4}^{119}}{{-}^{119}}{{C}_{1}}{{4}^{118}}\] \[{{+}^{119}}{{C}_{2}}{{4}^{117}}\,-\,\,119{{C}_{3}}{{4}^{116}}+......+(-1)\] \[\therefore \,\,\,\,{{3}^{119}}=4k-1\] \[\therefore \,\,\,{{3}^{120}}=\,\,12k\,-3=12\,(k-1)+9\] \[\therefore \] the required remainder is 9.
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