The difference between a number and the sum of its digits is divisible by 9.

## Solution

The difference between a number and the sum of its digits is divisible by 9. Therefore, all numbers we obtained were divisible by 9 (except, perhaps, the initial one). Let's go from the end. Zero can be obtained from any single-digit natural number after subtracting the sum of its digits. But among them, only 9 is divisible by 9. Therefore, on the penultimate step, we had the number 9. But 9 can only be obtained from one number divisible by 9 - from 18. And so on until we reach the number 81. Here the path splits - 81 can be obtained both from 90 and from 99. Let's take one more step back (now divisibility by 9 is no longer important!) - 90 cannot be obtained from any number, but for 99 there are 10 possible predecessors: $100, 101, 102, \ldots, 109$.

## Answer

Any number from 100 to 109.
Spivak A.V.

A natural number had the sum of its digits subtracted from it, then from the resulting number, the sum of its (resulting number's) digits was subtracted, and so on. After eleven such subtractions, zero was obtained. From which number did they start?