Power of 2^10 Minus Power of 10^3 is Divisible by 24/Proof 2
Jump to navigation
Jump to search
Theorem
Let $n \in \Z_{\ge 0}$ be a non-negative integer.
Then $2^{10 n} - 10^{3 n}$ is divisible by $24$.
That is:
- $2^{10 n} - 10^{3 n} \equiv 0 \pmod {24}$
Proof
For $n = 0$ both powers are $1$, and $1 - 1 = 0$ is divisible by $24$.
For $n > 1$:
\(\ds 2^{10 n} - 10^{3 n}\) | \(=\) | \(\ds 2^{3 n} \paren {2^{7 n} - 5^{3 n} }\) | ||||||||||||
\(\ds \) | \(\equiv\) | \(\ds 0\) | \(\ds \pmod 8\) | because $2^3 \divides 2^{3 n}$ | ||||||||||
\(\ds 2^{10 n} - 10^{3 n}\) | \(\equiv\) | \(\ds \paren {-1}^{10 n} - 1^{3 n}\) | \(\ds \pmod 3\) | Congruence of Powers | ||||||||||
\(\ds \) | \(\equiv\) | \(\ds 1 - 1\) | \(\ds \pmod 3\) | |||||||||||
\(\ds \) | \(\equiv\) | \(\ds 0\) | \(\ds \pmod 3\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds 2^{10 n} - 10^{3 n}\) | \(\equiv\) | \(\ds 0\) | \(\ds \pmod {24}\) | Chinese Remainder Theorem |
$\blacksquare$