Question #96785
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.
1
Expert's answer
2019-10-28T10:02:00-0400

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

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

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

Then a=19k2ba=19k-2b and hence


10a+b=10(19k2b)+b=190k19b=19(10k1)10a+b=10(19k-2b)+b=190k-19b=19(10k-1)

is a multiple of 19.19.

The condition is necessary too. Suppose 10a+b=19n10a+b=19n for some integer n.n. Then b=19n10ab=19n-10a and hence


a+2b=a+2(19n10a)=38n19a=19(2na)a+2b=a+2(19n-10a)=38n-19a=19(2n-a)

is a multiple of 19.19.


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