Power of 2^10 Minus Power of 10^3 is Divisible by 24/Proof 2

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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

$\blacksquare$