A Number That Counts Itself

Fill ten boxes, labelled $0, 1, 2, \dots, 9$, with single digits so that the digit in box $k$ says how many times $k$ appears in the ten-digit string you have written. So if box $3$ holds a $2$, the whole number must contain exactly two $3$s.

Solution

There is exactly one such number: $$6210001000.$$ Reading it against its own boxes: box $0$ holds $6$, and the number does contain six $0$s; box $1$ holds $2$, and there are two $1$s; box $2$ holds $1$, and there is a single $2$; box $6$ holds $1$, and there is a single $6$; every other box holds $0$, and those digits indeed do not appear.

A quick way to corner it is to notice that the ten box-entries must add up to $10$, because together they count each of the number’s ten digits exactly once. Here $6 + 2 + 1 + 1 = 10$, as it must. That balance condition, with the requirement that the entries describe themselves, leaves $6210001000$ as the only possibility.