Digital roots, and what survives repeated digit-summing

0:00Here is a six-digit number. Four hundred and eighty-nine thousand, seven hundred and ten. Add its digits. Four, eight, nine, seven, one, zero. That comes to twenty-nine. Twenty-nine is not a single digit, so do it again. Two and nine make eleven. Eleven is not a single digit either. One and one make two. Two. That is one digit, so you stop, and two is called the digital root of the number you started with.

0:32Three passes. And the question worth asking is not how to do it. It is what on earth that two is telling you. First, why does it stop at all? Nothing in the instruction says it has to. It stops because every pass makes the number strictly smaller. A six-digit number is at least a hundred thousand. The most its six digits can possibly add to is fifty-four. So one pass drops it from six digits to two, and the pass after that has almost nothing left to work with.

1:09Across nearly three thousand numbers, every pass came out smaller than the value before it, and every single number landed on one digit within three passes. Something that always shrinks and always stays positive has to stop. It is not lucky. It is forced. Now the important part. A pass throws away almost everything. It loses the size of the number. It loses how many digits it had. And it loses the order they were written in.

1:44Shuffle the digits of our number and you get seven hundred and twenty different numbers, from fourteen thousand seven hundred and eighty-nine up to nine hundred and eighty-seven thousand four hundred and ten. Every single one of them has digital root two. So whatever survives all this, it is not size, and it is not order. Let us measure what one pass actually removes. Our number was four hundred and eighty-nine thousand seven hundred and ten, and its digits added to twenty-nine. The pass removed four hundred and eighty-nine thousand six hundred and eighty-one.

2:27That amount is not arbitrary. It is fifty-four thousand four hundred and nine nines, exactly. And it is always like that. Across the sweep, every pass on every number removed an exact multiple of nine. The reason is how place value is built. Ten is one past a nine. A hundred is one past ninety-nine. A thousand is one past nine hundred and ninety-nine. So every digit contributes a run of nines plus one loose copy of itself. Adding the digits keeps the loose copies and throws the runs of nines away.

3:09And throwing away a multiple of nine cannot change how far a number sits past a multiple of nine. That is the one thing that survives. Every pass leaves it exactly where it was. So the single digit you finish on is carrying that, and nothing else. Which makes the digital root a report on division by nine. Our number leaves two when you divide it by nine. Its digital root is two.

3:39Across those same three thousand numbers, the root and the remainder agree two thousand six hundred and sixty-six times. Two thousand six hundred and sixty-six out of two thousand nine hundred and ninety-nine. So they disagree three hundred and thirty-three times. That is not a rounding error and it is not a rare exception. It is one case in nine. And the numbers where they disagree are exactly the multiples of nine.

4:11Take any multiple of nine and run the procedure. You always land on nine. But a multiple of nine leaves nothing over when you divide by nine. The remainder is zero. So which is it? Nine, or zero? Both, and there is no conflict once you see why. Zero and nine are the same position past a multiple of nine. That position is the multiple itself. The procedure simply cannot hand you zero. It only ever adds positive digits together, so it has no way of arriving there.

4:50Nine is the name this procedure gives that position. Nothing is broken, and nothing needs memorising. Now watch what happens along a run of numbers. Write twelve consecutive numbers starting from one, and take the root of each. One, two, three, four, five, six, seven, eight, nine. And then one, two, three again. Nine values, and then it repeats. Start anywhere and the same thing happens. Starting from our six-digit number the roots run two, three, four, five, six, seven, eight, nine, one, two, three, four.

5:30It is a ring of nine positions, and counting upwards walks you round it one step at a time. Different runs go round that ring at different speeds. Take consecutive multiples of three. Their roots go three, six, nine, three, six, nine. Three values, repeating. Multiples of six do the same thing. Six, three, nine, six, three, nine. Also three values. But multiples of four go four, eight, three, seven, two, six, one, five, nine.

6:06All nine values, one visit each, before it comes back round. Three and six share a factor with nine and four does not, and that one difference is what decides whether a run visits three positions or all nine of them. Adding is easier still, once you think in positions rather than in numbers. Ten is one past a multiple of nine. So adding ten moves the root on by one.

6:37A root of five becomes six. A root of nine becomes one, because the ring wraps round. That wrap is the part people drop. There is no tenth position to move to. Eleven is two past a multiple of nine, so adding eleven moves the root on by two, wrapping in exactly the same way. Start anywhere and keep adding eleven, and the roots step round all nine positions, two at a time.

7:07Across the sweep, both of those held on every single number. This also lets you read a root off an expression without knowing what the letters stand for. Take nine a, plus thirty-six b, plus thirteen. Nine a is a multiple of nine whatever a is. Thirty-six b is a multiple of nine whatever b is. Both of them sit at the position that means nothing left over, so neither one moves the root at all.

7:39Only the thirteen is left, and thirteen has digital root four. Try any pair of values you like. The answer comes out four every single time. Here is a riddle, described entirely by conditions. Every one of its digits is odd, and as small as an odd digit is allowed to be. And three separate things about it all come to the same value. How many digits it has, what its digits add up to, and its digital root.

8:12That value is the largest odd single digit. Nine. The first condition pins every digit to one, because one is the smallest odd digit there is. So it has nine digits, and all of them are ones. They add to nine, and nine has digital root nine. Nine ones in a row. Every condition lands on nine at once, and no other number does it. One last thing, and it is about a thousand years old.

8:44Aryabhata the second described this procedure, and it was used to check arithmetic before anyone had a machine to do it. Here is how. Multiply two numbers. Then take the root of each, multiply those two roots, and take the root of that. If your answer is right, the two must match. Over fourteen hundred and forty-four products, they matched every time. So a mismatch proves the calculation is wrong. That is genuinely useful and costs almost nothing.

9:18But a match proves much less than it looks like it proves. It only shows the two sides agree on their position past a multiple of nine, and an answer that is wrong by exactly nine agrees perfectly. Wrong by one gets caught. Wrong by nine walks straight through. Know which of those the check is telling you.