Remainders Quiz

$5 \\equiv -1 \\pmod{6}$. $5^{10} \\equiv (-1)^{10} = 1 \\pmod{6}$. Remainder = 1.

Cycle of $2^n \\pmod{5}$: $2, 4, 3, 1, 2, 4, ...$ (period = 4). $20 \\div 4 = 5$ remainder 0. So $2^{20} \\equiv 1 \\pmod{5}$.

By Fermat's Little Theorem: $a^{p-1} \\equiv 1 \\pmod{p}$ when $p$ is prime. $7^{52} \\equiv 1 \\pmod{53}$.

$3 \equiv -1 \pmod{4}$. $3^{100} \equiv (-1)^{100} = 1 \pmod{4}$. Remainder = 1.

$\varphi(7) = 6$. $3^6 \equiv 1 \pmod{7}$. $100 = 6 \times 16 + 4$. $3^{100} \equiv 3^4 = 81 \equiv 4 \pmod{7}$.

$\varphi(9) = 6$. $2^6 \equiv 1 \pmod{9}$. $256 = 6 \times 42 + 4$. $2^{256} \equiv 2^4 = 16 \equiv 7 \pmod{9}$.

$7 \equiv -1 \pmod{8}$. $7^{222} \equiv (-1)^{222} = 1 \pmod{8}$. Remainder = 1.

$17 \equiv 1 \pmod{16}$. $17^{123} \equiv 1^{123} = 1 \pmod{16}$. Remainder = 1.

$11^{99} = (10+1)^{99} = 100k + 99 \times 10 + 1 = 100k + 991$. Last two digits = 91.

Cyclicity of 9: $9^1=9$, $9^2=81\to1$, period 2. 2025 is odd, so units digit = 9.