The above number is over an octillion, yet I know that it is exactly divisible by 9. How do I know this? Easy. When you add the digits together, starting with the first 4 and then each group of 3 after that, you get 9+9+9+… And by random experiments I have conducted with calculators (and something I once had in grade school math IIRC), I know that when adding digits together, and you get all 9s, the whole number itself must be divisible by 9. (Another interesting application of this technique seems to happen when you add the digits of a number and get 3, thus showing it must be perfectly divisible by 3. E.g., 12, 201, 102, etc. are all evenly divisible by 3…)
Anyways, this is not MPSIMS, and my thread was not intended for MPSIMS. I simply want to know if there is a mathematical term for this: adding digits of a number together. And if there is a mathematical word or phrase for it, what is it? As you can see, I have known of this phenomenon since I was a child. But I never knew (or perhaps forgot) what the technical name for it is.
I’ve also only heard “digit sum.” Or the technical term “adding up the digits.”
Let me see if I can give a better proof for what you’re saying than “random experiments with calculators.”
Suppose you have a number X who’s digits are x0, x1, x2, . . . where xn is the digit in the 10^n place. (e.g. 145 has x0 = 5, x1 = 4, x2 = 1, and all others equal to zero.)
We can write X = x0 * 10^0 + x1 * 10^1 + x2 * 10^2 + . . .
10^n can be written as 1 + (an n-digit number made of only nines). e.g. 100 = 1 + 99
Rearranging this, we have X = A + B, where
A = x0 + x1 + x2 + x3 + . . .
B = x19 + x299 + x3*999 + . . .
Clearly B is divisible by 9 (since 9 = 19, 99 = 11 9, etc. and thus B is a sum of numbers divisible by 9). Thus, if A is divisible by 9 as well then we know X is divisible by 9 (since it’s a sum of numbers divisible by 9).
But A is just the sum of the digits, so this proves that if the sum of the digits is divisible by 9, then the number itself is divisible by 9 as well.
(Note that B is also divisible by 3, so if A is divisible by 3 then X must be as well.)
Yup, go with that… we had a Knowledge Master question with this a while back and we all stared at the screen like … We went and looked it up later lol.
I know that in his scientific american articles, Martin Gardner has used the term ‘casting out nines’ in reference to this procedure, or maybe more properly its result. I still like the term though and have used it to describe the process because it sounds so cool and old-fashioned mathematicky.
… We went and looked it up later lol.