this post was submitted on 17 Jul 2024
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Showerthoughts
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You can only use this method to check if the number can be divided by 3.
It works for 9, too.
If you're looking for a proof:
Our base 10 system represents numbers by having little multipliers in front of each power of 10. So a number like 1234 is 1 x 10^3 + 2 x 10^2 + 3 x 10^1 + 4 x 10^0 .
Note that 10 is just (3 x 3) + 1. So for any 2 digit number, you're looking at the first digit times (9 + 1), plus the second digit. Or:
(9 times the first digit) + (the first digit) + (the second digit).
Well we know that 9 times the first digit is definitely divisible by both 3 and 9. And we know that adding two divisible-by-n numbers is also divisible by n.
So we can ignore that first term (9 x first digit), and just look to whether first digit plus second digit is divisible. If it is, then you know that the original big number is divisible.
And when you extend this concept out to 3, 4, or more digit numbers, you see that it holds for every power of 10, and thus, every possible length of number. For both 9 and 3.