• # question_answer Find the remainder when $({{11}^{{{17}^{15}}}}+{{13}^{{{11}^{15}}}})$ is divided by 7. A)  0                     B)         1                     C)  2                     D)         3

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.