A) 3
B) 7
C) 9
D) 11
Correct Answer: C
Solution :
[c] \[{{27}^{40}}={{3}^{120}}\] \[{{3}^{119}}={{(4-1)}^{119}}={{\,}^{119}}{{C}_{0}}{{4}^{119}}-{{\,}^{119}}{{C}_{1}}{{4}^{118}}\] \[{{+}^{119}}{{C}_{2}}{{4}^{117}}-{{\,}^{119}}{{C}_{3}}{{4}^{116}}+...+(-1)\] \[\therefore {{3}^{119}}=4k-1\] \[\therefore {{3}^{120}}=12k-3=12(k-1)+9\] \[\therefore \] The required remainder is 9.
You need to login to perform this action.
You will be redirected in
3 sec
