Why the same digit sum also settles divisibility by 3

0:00You already have a way to test a number for nine. Add up its digits. Take four thousand four hundred and forty-four. Its digits add to sixteen, which is no multiple of nine. So the number is not one either. Now a different question about that same number. Is it a multiple of three? You could divide. Or you could look again at the sixteen you have already got. Sixteen is no multiple of three either. And that is the answer.

0:32One addition, asked twice. Not two rules - one rule, asked a second question. But that is worth nothing until you know why. Go back to the nines argument, and watch for where nine did some work. Take eighty-seven. Eight tens and seven ones. Rewrite each ten as nine plus one. Eight tens become eight nines plus eight ones. So eighty-seven splits in two. Eight nines, seventy-two. And eight plus seven, fifteen.

1:06Call the first part the heap, the second the loose digits. The loose part is exactly the digit total. Every number splits this way. A thousand splits into nine hundred and ninety-nine, and one. And here is the line to watch. The heap is always a multiple of nine. That is the only line where nine appears at all. So ask the obvious question. Is that heap a multiple of three too?

1:37The heap is built out of nines, ninety-nines, nine hundred and ninety-nines. Nine is three threes. Ninety-nine is three thirty-threes. Nine hundred and ninety-nine is three three-hundred-and-thirty-threes. Every run of nines is three times a run of threes, however long. So the heap is a multiple of nine, and therefore a multiple of three. That is no new fact about heaps. It is one fact about nine - whatever nine goes into, three goes into too.

2:14A second way to see this tells you which divisors can ever get such a test. Divide ten by three. Three threes, and one over. Divide a hundred by three. Thirty-three threes, and one over. A thousand? Three hundred and thirty-three threes, and one over. Every power of ten leaves one. Not usually - every one of the first thirty places, and every one of the next thirty too. That is what makes digits addable. If a place is worth one more than a multiple of three, seven of that place is worth seven more.

2:54Compare seven. There the powers of ten leave six different things, with no pattern to add up. Under eleven they leave two, flipping back and forth. Now run the split once more, with three in place of nine. Any number is a heap plus its digit total, and the heap is a multiple of three. So when you ask whether three divides the number, the heap is settled already. It contributes nothing.

3:25Everything rests on the loose part, which is the digit total. Eighty-seven again. Seventy-two and fifteen. Seventy-two is a multiple of three, so it drops out. Fifteen is a multiple of three. Therefore eighty-seven is. And notice what we did not do. We never divided eighty-seven. We divided fifteen. That is what a test is - a hard question swapped for an easy one with the same answer. So which divisors get a test like this? Not many, and there is a reason.

4:02Take the first twenty divisors and ask of each whether the digit total always leaves the same as the number does. Three of them pass. One, three and nine. Take the next twenty and none of them pass at all. And the three that work are exactly the three that go into nine. That is no coincidence, and no list to memorise. It falls straight out of the heap. The heap is a multiple of nine. If your divisor goes into nine, the heap is a multiple of your divisor too, and it drops out.

4:40This rule usually gets said one way round. If the digits add to a multiple of three, the number is a multiple of three. True. But that is half of what is true, and the missing half is the one you use most. The other direction says: if the number is a multiple of three, then its digits add to a multiple of three. Sweep both halves over three thousand numbers. Both hold. Sweep again four hundred thousand further up. Both still hold.

5:13Ask the same two halves about seven and both fail, so the sweep can say no. Why does the missing half matter? Because it is what lets the test refuse. With one direction only, a total of sixteen would tell you nothing. With both, sixteen is a real no. Here is where three and nine stop behaving alike. Every multiple of nine is a multiple of three - nine is three threes.

5:42The reverse is not true, and you can see the difference by asking about nine on one side and three on the other. This way round: a multiple of nine - does its digit total divide by three? Always. The other way round: a digit total divides by three - is the number a multiple of nine? No. Swap the two divisors over and the answers swap too. One half survives, and which one depends on how you asked.

6:14That lopsidedness is not a flaw. It is the test reporting that three goes into nine and nine does not go into three. Three numbers to hold on to. Fifteen, thirty-three, eighty-seven. Fifteen adds to six, thirty-three adds to six, eighty-seven adds to fifteen. All three are multiples of three. None is a multiple of nine. Watch out for a tempting wrong reason. It is not that the totals stop short of nine.

6:48Eighty-seven's total is fifteen, which is past nine. Add it again and you get six. Six is what all three really share. Six is a multiple of three and not of nine. How common is this? In nine hundred numbers, three hundred are multiples of three, one hundred of nine. Which leaves two hundred that are one and not the other - twice the multiples of nine. This is the ordinary case.

7:19So one digit total answers two questions, at two levels of detail. Add the digits, ask about three, then ask the same total about nine. Four ways two yes-or-no answers could come out. Only three ever happen. Yes to both. No to both. Yes to three, no to nine. The fourth never appears - nothing is a multiple of nine without being one of three. And here is a belief worth killing. Divisible by three twice over is not divisible by nine.

7:54A multiple of six times a multiple of three is always a multiple of nine. Sixteen hundred pairs, no exception. Push one step on and it stops. Ask whether that product is a multiple of twenty-seven, and only eight hundred and seventy-one of the sixteen hundred are. This test earns its keep on divisors that are not three or nine. Six is two threes. Test for two by the last digit, for three by the digit total. If both say yes, six divides it.

8:32Eighteen is two nines. Twenty-four is three eights. Thirty-six is four nines. Each is a digit test plus one easy other, and no division at all. But the parts must share no factor, and this is where people come unstuck. Twenty-four is also four sixes. So test for four, test for six, conclude twenty-four? It does not work. It gets the answer wrong on a hundred and twenty-five of the first three thousand numbers. The smallest is twelve.

9:07Four and six share a factor of two. You counted that two twice, and the number only had it once. Two last jobs, to see the test do what you would not want to do by hand. Take the six-digit number four eight, blank, two three, blank. Which digits make it a multiple of eighteen? Eighteen is two nines, so the last digit must be even and the total must reach a multiple of nine. Five pairs work.

9:39Drop the even condition, ask only for a multiple of nine, and there are eleven. Second. Between forty-five thousand and forty-seven thousand, how many multiples of thirty-six are there? Fifty-six. The first is forty-five thousand itself, the last is forty-six thousand nine hundred and eighty. Being asked for five of them is a choice, not a search. That is what one addition bought. Not a trick for nine and another for three, but one argument that mentioned neither.

10:15It only ever needed ten to be one past a multiple of your divisor. Nine is that multiple, and three is hiding inside it.