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Any number where the individual digits add up to a number divisible by '3' is divisible by 3.
51 = 5+1 = 6, which is divisible by three.
Try it, you'll see it always works.
There are tricks like that for a lot of numbers. For 7, chop off the last digit, double it and add it to what's left. Repeat as required. If the result is divisible by 7 then the original number was. eg: 356 -> 35+12=47 not db7. 357 =>35+14 both db7 so we don't even need to do the add.
Clever.
Girl.
double it and
add it tosubtract it from what’s left14: 1 + 8 = 7. Not db7
You clearly mean:
14: 1 + 8 = 9 (not db 7)
Someone else in this thread correctly stated:
"Chop of last digit, double it and subtract from what is left"
14: 1 - 8 = -7. (dB 7!)
Math is awesome, I didn't know this trick!
Yes, that's what I meant. The subtract rule works.
One of the reasons why I love the number 3. There are other neat digit sum tricks, see for example for the numbers 1 to 30 here: https://en.m.wikipedia.org/wiki/Divisibility_rule
They didn't teach stuff like this in school, which is silly. This is the kind of thing that a kid would eat up. It's like they wanted to make sure people hated math.
My experience of maths in high school was being taught a trick or method to solve a really specific type of problem every week. Sometimes the method would build off something we'd learnt the previous week.
The whole thing was bottom-up learning where you get given piecemeal nuggets of information but never see the big picture. They completely lost me at around the age of 15. I eventually came back to maths later in life after studying formal logic in my philosophy undergrad degree.
I guess I was one of the lucky few who learned this in elementary school. And later again.
I first saw the nine times finger trick in the movie 'Stand And Deliver.' I remember seeing it written out on a blackboard at some point, but never the finger trick.
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I knew that worked with 9. Hmm, does it work with 6?
Doesn’t look like it.
Technically it does work for 6, more literally, still aiming for 3, not 6. That's half of it, if the starting number is even and divisible by 3 then it is also divisible by 6.