Also, any number whose digits sum to a multiple of 3 is divisible by 3. For 51, 5+1=6, and 6 is a multiple of 3, so 51 can be cleanly divided by 3.
Same with 9. There are rules for every number at least through 13 that I once knew…
I only know rules for 2 (even number), 3 (digits sum to 3), 4 (last two digits are divisible by 4), 5 (ends in 5 or 0), 6 (if it satisfies the rules for both 3 and 2), 9 (digits sum to 9), and 10 (ends in 0).
I don’t know of one for 7, 8 or 13. 11 has a limited goofy one that involves seeing if the outer digits sum to the inner digits. 12 is divisible by both 3 and 4, so like 6, it has to satisfy both of those rules.
I’d forgotten this trick. It works for large numbers too.
122,300,223÷3 = 40,766, 741
1+2+2+3+2+2+3 = 15
wait till she finds out that 0.99999… 9’s to infinity is the same as 1
This one has always bothered me a bit because …999999 is the same as infinity, so when you’re “proving” this, you’re doing math using infinity as a real number which we all know it’s not.
You can also prove it a different way if you allow the use of the formula for finding the limit of the sum of a geometric series on a non-convergent series.
Sum(ar^n, n=0, inf) = a/(1-r)
So,
…999999
= 9 + 90 + 900 + 9000…
= 9x10^0 + 9x10^1 + 9x10^2 + 9x10^3…
= Sum(9x10^n, n=0, inf)
= 9/(1-10)
= -1
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.
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.
