[ Decimal number system ] [ Divisibility of numbers. General properties ]

Do there exist two consecutive natural numbers, the sum of the digits of each of which is divisible by 7?
If when transitioning from number $n$ to number $n+1$, $k$ digits are carried over, then the sum of the digits changes by $1 - 9k$.

## Solution

For example, the numbers 69999 and 70000.

## Answer

They exist.

What is the minimum sum of digits that a natural number divisible by $99$ can have?

## Hint

Use the divisibility rules for 9 and 11.

## Solution

An example with a digit sum of 18 is obvious - it is the number 99 itself. We will show that there are no numbers with a smaller digit sum that are divisible by 99. If a number $n$ is divisible by 99, then it is divisible by 9, so its digit sum must be divisible by 9. It remains to show that the digit sum cannot be 9.

By the divisibility rule for 11, the difference between the sums of the digits in the even and odd positions in the number $n$ must be divisible by 11. Let the sum of the digits in the even positions be $A$, and the sum of the digits in the odd positions be $B$, so $A - B$ is divisible by 11. Suppose the sum of the digits $A + B$ of the number $n$ is 9. Then $|A - B| \leq 9 < 11$. Moreover, $A - B \neq 0$, since one of the numbers $A, B$ is odd. This is a contradiction.

## Answer

18.

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