Divisibility by 5
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Theorem
An integer $N$ expressed in decimal notation is divisible by $5$ if and only if the units digit of $N$ is divisible by $5$.
That is:
- $N = \sqbrk {a_n \ldots a_2 a_1 a_0}_{10} = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$ is divisible by $5$
- $a_0$ is divisible by $5$.
Proof
Let $N$ be divisible by $5$.
Then:
\(\ds N\) | \(\equiv\) | \(\ds 0 \pmod 5\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \sum_{k \mathop = 0}^n a_k 10^k\) | \(\equiv\) | \(\ds 0 \pmod 5\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds a_0 + 10 \sum_{k \mathop = 1}^n a_k 10^{k - 1}\) | \(\equiv\) | \(\ds 0 \pmod 5\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds a_0\) | \(\equiv\) | \(\ds 0 \pmod 5\) | as $10 \equiv 0 \pmod 5$ |
$\blacksquare$
Sources
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): divisible
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): divisible