Who wants to multiply by 9, though? The version of the 13 test I've seen has you multiply by 4 and add instead:
4173 -> 417 + (3 x 4) = 429 -> 42 + (9 x 4) = 78, which is divisible by 13, so 4173 is.
Of course, these two tests work for the same reason. You're saying 10a + b is divisible by 13 iff a - 9b is, I'm saying 10a + b is divisible by 13 iff a + 4b is, and those differ by 13b.
But multiplying by 9 is so easy! You can check you’re right if (a) the first digit of the result is 1 less than the number you started with, and (b) the digits sum to 9
I was talking about the grand-parent comment's method:
> You can check you’re right if (a) the first digit of the result is 1 less than the number you started with
The first digit of the result is 2 ... And the number that you started with is also 2, which is not 1 less... I guess I am not smarter then a 5th grader
Wow! You definitely are smarter than me, the gp, at least! You’re right that my claim isn’t right in general. Hell, not even past 10, I can now see!
I’m curious, how did you find that counterexample? I kind of ‘overfitted’ my mental model to the first 10 integers. But as a statistician I’m only in the business of being approximately right...! Nice catch.
23x9=207 starts with 20 which is 3 less than 23... 24x9=216, and 21 is 3 less than 24. 11x9=99 and 9 is 2 less than 11, 12x9=108 and 10 is 2 less than 12. So 0-9 are offset by 1, 10-19 are offset by 2, 20-29 offset by 3, 30-49 offset by 4?
4173 -> 417 + (3 x 4) = 429 -> 42 + (9 x 4) = 78, which is divisible by 13, so 4173 is.
Of course, these two tests work for the same reason. You're saying 10a + b is divisible by 13 iff a - 9b is, I'm saying 10a + b is divisible by 13 iff a + 4b is, and those differ by 13b.