Remainder (2)

Number Theory Level 3

Find the remainder when $\Large \color{#D61F06} 2^{5^{9}} + 5^{9^{2}} + 9^{2^{5}}$ is divided by $\color{#D61F06}11$ .

5 7 6 8 4

1 solution

Chew-Seong Cheong
Jul 21, 2017

Since 2, 5, and 9 is each a coprime integer to 11, we can use Euler's theorem as follows:

$\begin{aligned} 2^{5^9} + 5^{9^2} + 9^{2^5} & \equiv 2^{5^9 \color{#3D99F6} \text{ mod }\phi (11)} + 5^{9^2 \color{#3D99F6} \text{ mod }\phi (11)} + 9^{2^5 \color{#3D99F6} \text{ mod }\phi (11)} \text{ (mod 11)} & \small \color{#3D99F6} \text{Euler's totient function }\phi(11) = 10 \\ & \equiv 2^{5^9 \color{#3D99F6} \text{ mod }10} + 5^{9^2 \color{#3D99F6} \text{ mod }10} + 9^{2^5 \color{#3D99F6} \text{ mod }10} \text{ (mod 11)} \\ & \equiv 2^5 + 5^{81 \text{ mod }10} + 9^{32 \text{ mod }10} \text{ (mod 11)} \\ & \equiv 32 + 5^1 + 9^2 \text{ (mod 11)} \\ & \equiv 10 + 5 + 4 \text{ (mod 11)} \\ & \equiv \boxed{8} \text{ (mod 11)} \end{aligned}$

Thank you so much. Nice and logical.

Hana Wehbi - 3 years, 10 months ago

Remainder (2)

1 solution

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