- #1
Unicyclist
- 42
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Okay, this may be a silly subject for a thread, but do you know that trick where you check whether a number is divisible by 3? The one where you add all the digits of that number, then add all the digits of the number you get as a result and so on, until you're left with a single-digit result. If the result is 3, 6 or 9, the number is divisible by 3.
Example 1: 4182
4 + 1 + 8 + 2 = 15
1 + 5 = 6 => 4182 is divisible by 3
Example 2: 9758
9 + 7 + 5 + 8 = 29
2 + 9 = 11
1 + 1 = 2 => 9758 is not divisible by 3
Does anybody know why this works and how you can prove it, other than by trial and error? I've always considered this a rather neat trick, but could never figure out how it works.
p.s. Well, yeah it's divisible, but 3 is not a factor of the number, so you won't get an integer as your result.
Example 1: 4182
4 + 1 + 8 + 2 = 15
1 + 5 = 6 => 4182 is divisible by 3
Example 2: 9758
9 + 7 + 5 + 8 = 29
2 + 9 = 11
1 + 1 = 2 => 9758 is not divisible by 3
Does anybody know why this works and how you can prove it, other than by trial and error? I've always considered this a rather neat trick, but could never figure out how it works.
p.s. Well, yeah it's divisible, but 3 is not a factor of the number, so you won't get an integer as your result.