The digit-sum test for 9, and the algebra underneath it

0:00Here is a number. Four thousand and seventy-five. What is the seven doing there? It is not worth seven. It is worth seventy. Every digit is worth what it is because of where it stands, and if you write that out you get four thousands, no hundreds, seven tens, and five. Add those back up and you have the number again. That is not a trick. That is what the number is.

0:27And that expansion buys you something immediately. Is four thousand and seventy-five a multiple of ten? Look at the pieces. Thousands are multiples of ten. Hundreds are multiples of ten. Tens are multiples of ten. Everything above the units place is already a multiple of ten, so none of it can tip the answer either way. Only the last digit can. It is five. So no. Now a harder question. Is ninety-nine thousand and nine a multiple of nine?

1:05You can settle that one without dividing, and here is the tempting way to do it. It is built out of nines and noughts. Nine ten-thousands, nine thousands, no hundreds, no tens, and nine. Every piece with a nine in it is nine times something, and the noughts contribute nothing at all. So yes. And its digits add to twenty-seven, which you may already suspect is not a coincidence. The same argument works for nine hundred and ninety-nine, for nine hundred and nine, for nine hundred, for ninety, for nine hundred and ninety.

1:45Every one of them. Which feels like a rule. It is not a rule. It is a special case, and it is about to mislead you. Because here is the obvious next thought. Look at the last digit, the way we just did for ten. Ninety-nine ends in nine, and ninety-nine is a multiple of nine. A hundred and nine also ends in nine. It is not. Divide it and you get twelve, with one left over.

2:16So the last digit settles nothing here. And that is not bad luck, it is structural. Ten's test worked because every place value above the units was already a multiple of ten. Nine has no such luck. Ten is not a multiple of nine, and neither is a hundred. So the higher places do contribute something, and we have to find out exactly what. Start small. Divide ten by nine.

2:47One, with one left over. Twenty? Two nines are eighteen, so two left over. Thirty leaves three. Forty leaves four. There is the pattern. The leftover is simply the number of tens. And it has to be, because every ten is one past a nine. So forty is four nines, and then four stragglers, one from each ten. Now ask the same thing of hundreds. A hundred divided by nine is eleven, with one left over.

3:20Two hundred leaves two. Three hundred leaves three. Four hundred leaves four. Same pattern. The leftover is the count of hundreds. And the reason is the same shape as before. A hundred is ninety-nine plus one, and ninety-nine is eleven nines. So four hundred is a large heap of nines, and four stragglers. Whatever the place value, each copy of it contributes exactly one. Which means the leftover from any place is just the digit standing in it.

3:57Let us put that to work on a real number. Four hundred and twenty-seven. Split it into its places. Four hundred, twenty, and seven. Four hundreds contribute four. Two tens contribute two. Seven units contribute seven. Add the stragglers up. Four and two and seven is thirteen. But thirteen cannot be a leftover under nine. There is another nine hiding inside it. Take that nine out and four remains. So four hundred and twenty-seven leaves four.

4:32And look at what we actually did to get there. We added the digits. So why does adding the digits work? Line the place values up. One is nought plus one. Ten is nine plus one. A hundred is ninety-nine plus one. A thousand is nine hundred and ninety-nine plus one. Every single place value is one past a run of nines. And a run of nines is always a multiple of nine. Nine is. Ninety-nine is eleven nines. Nine hundred and ninety-nine is a hundred and eleven of them.

5:12That is not a coincidence about small numbers, and it does not run out. Ten to any power at all, however high you go, is one past a run of nines. That single identity is the entire engine. Now watch what happens when you feed that in. Take any number and write it in its places. Then replace every place value by its run of nines, plus one. Multiply that out and the number falls into two heaps.

5:44The first heap is every digit multiplied by a run of nines. That heap is visibly a multiple of nine. It is made of nothing else, so it cannot be anything else. Which means it contributes nothing whatsoever to the leftover. The second heap is one loose copy of each digit. And one loose copy of each digit, added up, is the digit sum. Do it on a number. Seven thousand three hundred and nine.

6:16Seven thousands, three hundreds, no tens, nine units. Replace each place value. Seven times nine hundred and ninety-nine, plus seven. Three times ninety-nine, plus three. And nine. The heap of nines comes to seven thousand two hundred and ninety. The loose part is seven, plus three, plus nought, plus nine. That is nineteen. Seven thousand two hundred and ninety, plus nineteen, is the number we started with. The first part leaves nothing at all. So the number leaves exactly what nineteen leaves.

6:55The number and its digit sum sit the same distance past the nearest multiple of nine. Always. That is the whole theorem. Nineteen is still too big to be a leftover, so do the step again. One plus nine is ten. Still too big. One plus nought is one. So seven thousand three hundred and nine leaves one. Divide it and check, if you like. And that gives you a rule you may have met. Keep adding the digits until one digit is left, and that digit is the leftover.

7:32It is very nearly right. There is exactly one place where it lies to you. Take nine hundred and ninety-nine. The digits add to twenty-seven, and twenty-seven adds to nine. But nine hundred and ninety-nine leaves nothing. It is a multiple of nine. So on every multiple of nine the rule says nine when the honest answer is nought. Which is worth knowing, because a multiple of nine is precisely what people use the rule to look for.

8:07Now, the test runs in both directions, and that is a much stronger claim than it sounds. A number is a multiple of nine exactly when its digit sum is. That is two statements wearing one coat. If the number is a multiple, the digits add to a multiple. And if the digits add to a multiple, the number is one. Plenty of tests only go one way. This one goes both, and so do the two negative versions — all four stand or fall together.

8:41Which is what lets you run it backwards. Here is a four-digit number. Three, one, something, five, and I am telling you it is a multiple of nine. Three and one and five make nine. So the missing digit has to leave the total a multiple of nine. Nought does that. And nine does it too. Two answers, not one. And that is the ordinary case, not a quirk of this particular number.

9:11One last thing, and it is the part that usually gets left out. Nothing in that argument was really about nine. The only fact we ever used was that ten is one past a multiple of nine. So ask which other divisors have that property, and the whole argument comes along unchanged. Ten is also one past a multiple of three. So the digit sum test works for three, for exactly the same reason, and nobody had to prove it twice.

9:45Sweep the first twenty divisors and precisely three of them work. One, three and nine. Not seven. Ten leaves three under seven, not one, and the argument collapses at its very first step. So this was never a fact about the number nine. It is a fact about how we write numbers down — about ten — and nine is simply the number that ten happens to sit one past.

10:13Write your numbers in a different base and you would be adding digits to test something else entirely.