JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Binomial Theorem

We compute $7^{103}\mathrm{mod}17$ using binomial expansion:


Expanding $(51-2{)}^{51}$ by the binomial theorem, all terms except the last are divisible by 51 (hence by 17), leaving remainder $(-2{)}^{51}$.

So the remainder is $7\times (-2{)}^{51}=-7\times 2^{51}$.

Now $2^{51}=8\times 2^{48}=8\times (2^4{)}^{12}=8\times 16^{12}$.

Since $16\equiv -1(\mathrm{mod}17)$, we have $16^{12}\equiv 1(\mathrm{mod}17)$.

Thus $2^{51}\equiv 8(\mathrm{mod}17)$, and the remainder is $-7\times 8=-56\equiv -56+68=12(\mathrm{mod}17)$.