Power of 2^10 Minus Power of 10^3 is Divisible by 24/Proof 1

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

\(\ds 2^{10 n}\)

\(=\)

\(\ds \paren {2^{10} }^n\)

Power of Power

\(\ds \)

\(=\)

\(\ds 1024^n\)

as $2^{10} = 1024$

\(\ds \)

\(=\)

\(\ds \paren {1000 + 24}^n\)

rewriting $1024$ as the sum of a power of $10$ and some integer

\(\ds \)

\(=\)

\(\ds \sum_{k \mathop = 0}^n 1000^{n - k} \, 24^k\)

Binomial Theorem

\(\text {(1)}: \quad\)

\(\ds \)

\(=\)

\(\ds 1000^n + \sum_{k \mathop = 1}^n 1000^{n - k} \, 24^k\)

extracting first term from summation

\(\ds \)

\(=\)

\(\ds 1000^n + 24 \paren {\sum_{k \mathop = 1}^n 1000^{n - k} \, 24^{k - 1} }\)

extracting $24$ as a divisor

\(\ds \leadsto \ \ \)

\(\ds 2^{10 n} - 1000^x\)

\(=\)

\(\ds 24 i\)

setting $i = \ds \sum_{k \mathop = 1}^n 1000^{n - k} \, 24^{k - 1}$

\(\ds \leadsto \ \ \)

\(\ds 2^{10 n} - 10^{3 n}\)

\(=\)

\(\ds 24 i\)

rewriting $1000^n$ as a power of $10$

\(\ds \leadsto \ \ \)

\(\ds 2^{10 n} - 10^{3 n}\)

\(\equiv\)

\(\ds 0 \pmod {24}\)

Definition of Congruence Modulo Integer

$\blacksquare$