this post was submitted on 17 Jul 2024
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Divisible by 3. Easy to check since 7 + 1 + 7 = 15 which is divisible by 3.
Oh awesome that's a neat trick I've never seen before. How does that work? For a number like 700 for example, 7 + 0 + 0 = 7 but 700 is visible by 10.
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.
It works differently for each number. For 2, the last number has to be divisible by 2. For 3, the sum of the digits has to be divisible by 3 For 5, the number has to end with a 0 or a 5. For 7, it is kinda tricky. Take the last digit, double it, and subtract it from the numbers on the left. If the remainder is 0 or divisible by 7, the whole number is divisible by 7. For example 49: 9×2=18, 4-18=-14, -14/7=2 with remainder 0. For 700, 0×2=0, 70-0=70, 70/7=10 remainder 0.
This is usually specified for prime numbers, for non-prime number, you just do calculate the prime components of a number and combine the rules.
For example, divisibility by 15: it has to be divisible by 3 and 5. 1+5=6, 6/3=2 remainder 0. 15 ends with a 5. For number where with multiple same prime components the rules for these duplicate numbers have to apply multiple times. Like for 25, it has to end with a 5 or 0, and when dividing the number by 5, the result has to end with a 5 or a 0 aswell.
A programmer I know wrote a small paper about this
Back in the olden days before digital calculators, people had tricks like that to help them do mental math.
😂 I'm familiar with many of them because I liked math in school and went through all of primary and middle school without a digital calculator. This one is one I hadn't heard of before and, again because I like math, I'm interested in understanding 1) what the trick is and 2) mathematically how it works out