A number is a multiple of 3 if you can add up the digits and the sum is also a multiple of 3. But why does it work that way? And do similar phenomena appear in other base number systems?

Question

A number is a multiple of 3 if you can add up the digits and the sum is also a multiple of 3. But why does it work that way? And do similar phenomena appear in other base number systems?

In: 17

Answer 1

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Answer 2

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Answer 3

It’s also works for multiples of 9, which is one less than our base. 3 is the square root of 9, so also works.

This generalizes to every base n, for n-1, and also any ~~root~~ factor of (n-1). In base-6, the trick works for mutiples of 5. In base-17, it would work for multiples of 16, but also multiples of 2, 4 or 8.

Answer 4

Because if you have a number xyz it can also be written as 100x + 10y + z. The 100 and 10 multiplicators can be rewritten as 9+1 or 99+1. You can discard the 9 and 99 because they are multiples of 3 (and 9 that’s why it also works for 9) so xyz=100x + 10y + z = x+y+z +(99x + 9y)

Answer 5

It’s also works for multiples of 9, which is one less than our base. 3 is the square root of 9, so also works.

This generalizes to every base n, for n-1, and also any ~~root~~ factor of (n-1). In base-6, the trick works for mutiples of 5. In base-17, it would work for multiples of 16, but also multiples of 2, 4 or 8.

Answer 6

Because if you have a number xyz it can also be written as 100x + 10y + z. The 100 and 10 multiplicators can be rewritten as 9+1 or 99+1. You can discard the 9 and 99 because they are multiples of 3 (and 9 that’s why it also works for 9) so xyz=100x + 10y + z = x+y+z +(99x + 9y)