Answer to Question #96785 in Combinatorics | Number Theory for Dina Patel

Question

To check whether a number is divisible by 19 one can repeatedly add twice the last digit
to the rest of the number. For example:
13547→ 1354 + 7 ∗ 2 = 1368 → 136 + 8 ∗ 2 = 152 → 15 + 4 = 19
thus 13547 is divisible by 19 (indeed, 13547=19*713). Give a proof of this divisibility rule.

Answer 1

Suppose we have the integer number $10a+b,$ where $b$ is the digit of the units (this writing form is unique if we say that $b$ is a digit).

The rule says to consider $a+2b.$

Suppose this is a multiple of $19,$ namely $a+2b=19k,$ where $k$ is some integer number.

Then $a=19k-2b$ and hence

$$10a+b=10(19k-2b)+b=190k-19b=19(10k-1)$$

is a multiple of $19.$

The condition is necessary too. Suppose $10a+b=19n$ for some integer $n.$ Then $b=19n-10a$ and hence

$$a+2b=a+2(19n-10a)=38n-19a=19(2n-a)$$

is a multiple of $19.$