The remainder when 19^200+23^200 is divided by 49, is _______.

Question

The remainder when $19^{200}+23^{200}$ is divided by 49, is \_\_\_\_\_\_\_.

Answer: 29

Accepted Answer

$$(21+2)^{200}+(21-2)^{200}=2\left[2^{200}+\binom{200}{2}21^2\cdot2^{198}+\cdots\right]=49I_1+2\cdot2^{200}=49I_1+2^{201}$$

Now $2^{201}=8^{67}=(1+7)^{67}=49I_2+{}^{67}C_0+{}^{67}C_1\times7=49I_2+1+469=49I_2+49\times9+29$

So $19^{200}+23^{200}\equiv29\pmod{49}$.

Solutions · 1