The remainder, when 7^103 is divided by 23, is equal to:

Question

The remainder, when $7^{103}$ is divided by 23, is equal to:

Accepted Answer

$$7^{103}=7\cdot7^{102}=7\cdot343^{34}=7(345-2)^{34}=7\left(23\cdot15-2\right)^{34}$$

Expanding by the binomial theorem, all terms except the last are multiples of 23:

$$7^{103}=23K_1+7\cdot(-2)^{34}=23K_1+7\cdot2^{34}$$

Now $7\cdot2^{34}=7\cdot2^2\cdot2^{32}=28\cdot256^4=28(253+3)^4=28\big[23m+3^4\big]=23m'+28\times81$

$28\times81=2268=23\times98+14$

So $7^{103}\equiv14\pmod{23}$. Remainder = 14

Solutions · 1