# Every Tenth Power of Two Minus Every Third Power of Ten is Divisible By Three

## Theorem

Let $x \in \Z_{\ge 0}$ be a non-negative integer.

Then $2^{10 x} - 10^{3 x}$ is divisible by $3$.

That is:

$2^{10 x} - 10^{3 x} \equiv 0 \pmod 3$

## Proof

 $\ds 2^{10 x}$ $=$ $\ds \paren {2^{10} }^x$ Power of Power $\ds$ $=$ $\ds 1024^x$ as $2^{10} = 1024$ $\ds$ $=$ $\ds \paren {1000 + 24}^x$ rewriting $1024$ as the sum of a power of $10$ and some integer $\ds$ $=$ $\ds \sum_{k \mathop = 0}^n 1000^{x - k} \, 24^k$ Binomial Theorem $\text {(1)}: \quad$ $\ds$ $=$ $\ds 1000^x + \sum_{k \mathop = 1}^n 1000^{x - k} \, 24^k$ extracting first term from summation $\ds$ $=$ $\ds 1000^x + 24 \paren {\sum_{k \mathop = 1}^n 1000^{x - k} \, 24^{k - 1} }$ extracting $24$ as a divisor $\ds \leadsto \ \$ $\ds 2^{10 x} - 1000^x$ $=$ $\ds 24 i$ setting $i = \ds \sum_{k \mathop = 1}^n 1000^{x - k} \, 24^{k - 1}$ $\ds \leadsto \ \$ $\ds 2^{10 x} - 10^{3 x}$ $=$ $\ds 24 i$ rewriting $1000^x$ as a power of $10$ $\ds$ $=$ $\ds 3 k$ setting $k = 8 i$ $\ds \leadsto \ \$ $\ds 2^{10x} - 10^{3 x}$ $\equiv$ $\ds 0 \pmod 3$ Definition of Congruence Modulo Integer

$\blacksquare$

## Examples

Let us examine arguably the flagship example, the smallest number whose compliance is not obvious nor trivial:

$1 \, 048 \, 576$

This is equal to $2^{20}$, which is equal to $2^{10 \times 2}$, thus one of the valid powers of $2$.

We then subtract $10^{3 \times 2}$, or $10^6$:

$1 \, 048 \, 576 - 1 \, 000 \, 000 = 48 \, 576$

and we see that:

$48 \, 576 = 16 \, 192 \times 3$

satisfying the theorem.