Powers of 16 Modulo 20/Proof 2
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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 |
$\blacksquare$