Reaminder Problem 1

First note that $16^{53} = (14 + 2)^{53} \equiv 2^{53} \pmod{7}$ .

Now $2^{3} = 8 \equiv 1 \pmod{7}$ . Also, $53 = 17 * 3 + 2$ .

Thus $2^{53} \equiv 2^{2} \pmod{7} \equiv 4 \pmod{7}$ , i.e., the desired remainder is $\boxed{4}$ .

This may also be solved using Fermat's Little Theorem. As in Brian's solution, note that $16^{53}\equiv 2^{53}\pmod {7}$ .

Next, note that $2^{49}\equiv\left(2^{7}\right)^7\pmod 7$ . Thus, by FLT, we have $2^{49}\equiv 2\pmod 7$ .

Putting it all together, $2^{53}\equiv 2^{49} 2^4\equiv2^5\equiv 4 \pmod 7$

(16^1) /7 remainder is 2

(16^2) /7 remainder is 4

(16^3) /7 remainder is 1

(16^4) /7 remainder is 2 ............

53/3 remainder is 2

ans is 4

The remainder of : (when divided by 7) $16^{1}=2$ $16^{2}=4$ $16^{3}=1$ $16^{4}=2$ ....... This goes on repeating after this

53=51 + 2

Second term in the above pattern is 4

Thus, remainder of $16^{53}$ is 4