A) 0
B) 1
C) 2
D) 3
Correct Answer: B
Solution :
When 17" is divided by 6 \[\frac{{{(18-1)}^{15}}}{6},\] remainder = 5 So, \[{{17}^{15}}\] can be written as \[6k+5\]. \[\therefore \] \[\frac{{{11}^{{{17}^{15}}}}}{7}=\frac{{{11}^{6k+5}}}{7}=\frac{{{(7+4)}^{6k+5}}}{7}=\frac{6k+5}{7}\] \[=\frac{16\times {{({{4}^{3}})}^{2k+1}}}{7}=\frac{16\times {{(63+1)}^{2k+1}}}{7}\] \[\frac{{{13}^{{{11}^{15}}}}}{7}=\frac{{{(14-1)}^{\text{odd}}}}{7}\Rightarrow \] Remainder = 6 So, remainder when \[{{11}^{{{17}^{15}}+}}{{13}^{{{17}^{15}}}}\] is divided by 7 is 1.
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