The Suppressed Digit

Here is a trick to play on a friend. Ask them to do the following out of your sight.

1. Write down any whole number with more than five digits.

2. Add up its digits, and subtract that total from the number itself.

3. From the result, cross out any one digit that is not a zero.

4. Read out the digits that remain, in any order they please.

You then name the digit they crossed out. If they read out $2, 2, 3, 5, 9$, you would tell them at once that the suppressed digit was a $6$. How is it done?

Solution

The trick rests on a single fact about the number nine: a whole number and the sum of its digits leave the same remainder on division by nine. This holds because $10$, $100$, $1000$ and so on are each one more than a multiple of nine, so every digit contributes only itself to the remainder.

It follows that subtracting the digit sum from the number always leaves an exact multiple of nine. And the digits of a multiple of nine themselves add up to a multiple of nine.

So the digits of the result, all of them together, add up to a multiple of nine. The ones read out to you sum to $2 + 2 + 3 + 5 + 9 = 21$. The next multiple of nine above $21$ is $27$, so the suppressed digit must be $27 - 21 = 6$.

This also explains the rule against crossing out a zero. If the digits you hear already add up to a multiple of nine, the hidden digit is a $9$, not a $0$, since a suppressed $0$ and a suppressed $9$ could not be told apart. Barring zero removes the ambiguity.