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

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

${\left(51-2\right)}^{51}·7=7·\left({}^{51}C_0{51}^{51}·2^0-{}^{51}C_1{51}^{50}·2^1+.......-{}^{51}C_{51}{2}^{51}\right)$

Now $51$ is divisible by $17$, so when above equation is divided by $17$ we get, $7·\left(-2^{51}\right)$ as remainder,

Now again using binomial in $7·\left(-2^{51}\right)=-56{\left(2^4\right)}^{12}=-56{\left(17-1\right)}^{12}$ we get,

$-56{\left(17-1\right)}^{12}=-56\left({}^{12}C_0{17}^{12}·1^0-{}^{12}C_1{17}^{11}·1^1+.......+{}^{12}C_{12}{1}^{12}\right)$

Now dividing above equation by $17$ we get,

$\left(-56\times 1\right)÷17=-5$,

Now changing negative remainder to positive we get, $-5+17=12$,

Hence, when $7^{103}$ divided by $17$ gives $12$ as remainder.