Divisibility by 7

Theorem

An integer $X$ with $n$ digits ($X_0$ in the ones place, $X_1$ in the tens place, and so on) is divisible by $7$ if and only if $\displaystyle \sum_{i=0}^{n-1} (3^i X_i)$ is divisible by $7$.

Direct Proof

	\(\displaystyle \)	\(\displaystyle X\)	\(=\)	\(\displaystyle 10^0 X_0 + 10^1 X_1 + \cdots + 10^{n-1} X_{n-1}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{i=0}^{n-1} 10^i X_i\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{i=0}^{n-1} (10^i - 3^i + 3^i) X_i\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{i=0}^{n-1} ((10^i - 3^i) X_i) + \sum_{i=0}^{n-1} (3^i X_i)\)	\(\displaystyle \)

The first addend is always divisible by $7$ because $10^i - 3^i$ always produces a number divisible by $7$ from the difference of two powers. So $X$ will be divisible by $7$ if and only if the second addend is divisible by $7$.

$\blacksquare$