5 ms·

57 is 3 times 19. The standard divisibility rule for 3, 6 and 9 in base 10 is to sum the digits until you only have one left and check if it's one of those. He

57 is 3 times 19. The standard divisibility rule for 3, 6 and 9 in base 10 is to sum the digits until you only have one left and check if it's one of those. Here, 5+7=12, 1+2=3, so 57 is divisible by 3.

jeffbee 10mo ago
Ah I see what is meant by recursively here. Thanks!
squigz 10mo ago
Math is not my strong suit at all, so I probably won't grok this, but that kind of blows my mind, so I'm curious... how?! That works for any arbitrarily large number? Math is crazy!... still don't want to study it though!
- moring 10mo ago
  Yes. A number like 123456 = 1 * 100000 + 2 * 10000 + 3 * 1000 + 4 * 100 + 5 * 10 + 6 = 1 * (99999+1) + 2 * (9999+1) + 3 * (999+1) + 4 * (99+1) + 5 * (9+1) + 6 When checking whether it is a multiple of some k, you can add/subtract multiples of k without changing the result, and those 99...9 are multiples of both 3 and 9. So 123456 is a multiple of 3 (or 9) iff 1 * 1 + 2 * 1 + 3 * 1 + 4 * 1 + 5 * 1 + 6 * 1 = 1 + 2 + 3 + 4 + 5 + 6 is. Apply the same rule as often as you want -- that is, until you only have one digit left, because then it won't get simpler anymore.
- deleted 10mo ago
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- freehorse 10mo ago
  It is basically because $10 mod 3 == 1$ (as 10 = 3*3 + 1). So if you are in the ring modulo 3, where every number is equal to the remainder of its division by 3, the sum of the digits of the number in its decimal representation equals the number itself (modulo 3), because in that ring 10 is actually 1, so the 10s in the decimal sum become 1s. Ie if n_k is the kth digit of n, you have (mod 3) n == n_0 + n_1*10 + n_2*10^2 + ... == n_0 + n_1 + n_2 + ... Hence, n is divisible by 3 iff $n mod 3 == 0$ iff $(n_0 + n_1 + n_2 + ...) mod 3 == 0$. Of course, summing up the digits may not give you a 1-digit number, but it gives you a number that you know is divisible by 3 (if the original number is divisible by 3). So you can apply the same idea/process again, summing up the digits of that number, and get another number that is divisible by 3. Repeat until you end up with one digit (hence the recursion mentioned).

jeffbee 10mo ago

Ah I see what is meant by recursively here. Thanks!

squigz 10mo ago

Math is not my strong suit at all, so I probably won't grok this, but that kind of blows my mind, so I'm curious... how?! That works for any arbitrarily large number? Math is crazy!... still don't want to study it though!

moring 10mo ago
Yes. A number like 123456 = 1 * 100000 + 2 * 10000 + 3 * 1000 + 4 * 100 + 5 * 10 + 6 = 1 * (99999+1) + 2 * (9999+1) + 3 * (999+1) + 4 * (99+1) + 5 * (9+1) + 6 When checking whether it is a multiple of some k, you can add/subtract multiples of k without changing the result, and those 99...9 are multiples of both 3 and 9. So 123456 is a multiple of 3 (or 9) iff 1 * 1 + 2 * 1 + 3 * 1 + 4 * 1 + 5 * 1 + 6 * 1 = 1 + 2 + 3 + 4 + 5 + 6 is. Apply the same rule as often as you want -- that is, until you only have one digit left, because then it won't get simpler anymore.
deleted 10mo ago
[deleted]
freehorse 10mo ago
It is basically because $10 mod 3 == 1$ (as 10 = 3*3 + 1). So if you are in the ring modulo 3, where every number is equal to the remainder of its division by 3, the sum of the digits of the number in its decimal representation equals the number itself (modulo 3), because in that ring 10 is actually 1, so the 10s in the decimal sum become 1s. Ie if n_k is the kth digit of n, you have (mod 3) n == n_0 + n_1*10 + n_2*10^2 + ... == n_0 + n_1 + n_2 + ... Hence, n is divisible by 3 iff $n mod 3 == 0$ iff $(n_0 + n_1 + n_2 + ...) mod 3 == 0$. Of course, summing up the digits may not give you a 1-digit number, but it gives you a number that you know is divisible by 3 (if the original number is divisible by 3). So you can apply the same idea/process again, summing up the digits of that number, and get another number that is divisible by 3. Repeat until you end up with one digit (hence the recursion mentioned).