8.5. Each digit of the natural number $N$ is strictly greater than the one to its left. What is the sum of the digits of the number $9 N$?
Answer: 9.

Solution: Note that $9 N=10 N-N$. Let's perform this subtraction in a column. In the units place, there will be the difference between 10 and the last digit of the number $N$, in the tens place - the last and the second-to-last digit, decreased by 1. In all subsequent places, there will be the difference between two adjacent digits, as the smaller digit will always be subtracted from the larger one. Each digit will appear once as the minuend and once as the subtrahend. When finding the sum of the digits, all these minuends and subtrahends will cancel each other out. Only the aforementioned $10-1=9$ will remain.

Criteria: only the answer, the answer with verification on specific cases - 0 points, the scheme of proof by example -2 points, the transition $9 N=10 N-N-1$ point. These points $\mathrm{NE}$ are not summed.

## Solutions to the tasks of the first stage of the All-Siberian Open Olympiad for schoolchildren in mathematics 2014-15

## 9th grade

## Each task is worth 7 points