The remainder, when 7 ^ 103 is divided by 17, is

Using the binomial expansion in ${7}^{103} = {\left({7}^{2}\right)}^{51} \cdot 7 = {\left(51 - 2\right)}^{51} \cdot 7$ we get,

${\left(51 - 2\right)}^{51} \cdot 7 = 7 \cdot \left(C0 51 {51}^{51} \cdot {2}^{0} - C1 51 {51}^{50} \cdot {2}^{1} + . . . . . . . - C51 51 {2}^{51}\right)$

Now $51$ is divisible by $17$, so when above equation is divided by $17$ we get, $7 \cdot \left(- {2}^{51}\right) as remainder,</p><p>Now again using binomial$ in 7 \cdot \left(- {2}^{51}\right)$= - 56{\left({2}^{4}\right)}$^{12} = - 56{\left(17 - 1\right)}^{12} we get,

$- 56{\left(17 - 1\right)}$^{12} = - 56\left(C0 12 {17}^{12} \cdot {1}^{0} - C1 12 {17}^{11} \cdot {1}^{1} + . . . . . . . + C12 12 {1}^{12}\right)$</p><p>Now dividing above equation by$17$we get,</p><p>\left(- 56 \times 1\right)$ \div 17 = - 5$,</p><p>Now changing negative remainder to positive we get,$- 5 + 17 = 12$,</p><p>Hence, when${7}^{103}$divided by$17$gives$12$as remainder.</p></blockquote>$