Powers of 16 Modulo 20/Proof 2

Theorem

Let $n \in \Z_{> 0}$ be a strictly positive integer.

Then:

$16^n \equiv 16 \pmod {20}$

Proof

\(\ds 16\)

\(\equiv\)

\(\ds 16\)

\(\ds \pmod {20}\)

\(\ds \leadsto \ \ \)

\(\ds 16\)

\(\equiv\)

\(\ds 0\)

\(\ds \pmod 4\)

\(\, \ds \text {and} \, \)

\(\ds 16\)

\(\equiv\)

\(\ds 1\)

\(\ds \pmod 5\)

\(\ds \leadsto \ \ \)

\(\ds 16^n\)

\(\equiv\)

\(\ds 0\)

\(\ds \pmod 4\)

\(\, \ds \text {and} \, \)

\(\ds 16^n\)

\(\equiv\)

\(\ds 1\)

\(\ds \pmod 5\)

\(\ds \leadsto \ \ \)

\(\ds 16^n\)

\(\equiv\)

\(\ds 16\)

\(\ds \pmod {20}\)

Chinese Remainder Theorem

$\blacksquare$