Answer:

288

## Solution

Note that the number 288 has divisors $1,2,3,4,32,6,16,8,9$. Therefore, this number satisfies the condition of the problem. We will prove that there is no smaller number that satisfies the condition.

Indeed, since $\mathrm{N}$ must have a divisor with the sum of digits 9, $\mathrm{N}$ is divisible by 9.

Now consider the divisor $\mathrm{d}$ with the sum of digits $8. \mathrm{d}$ is not divisible by 3, so the numbers $\mathrm{d}$ and 9 are coprime, which means $\mathrm{N}$ is divisible by 9d. At the same time, if $\mathrm{d} \geq 32$, then $9 \mathrm{~d} \geq 288$, that is, $3 \mathrm{~N} \geq$ 288. Therefore, it remains to check $\mathrm{d}=26, \mathrm{~d}=17$ and $\mathrm{d}=8$.

If $\mathrm{d}=26$, then $9 \mathrm{~d}=234$. This number does not have a divisor with the sum of digits 5, and any number that is a multiple of it is greater than 288.

If $\mathrm{d}=17$, then $9 \mathrm{~d}=153$. This number does not have a divisor with the sum of digits 2, and any number that is a multiple of it is greater than 288.

If $\mathrm{d}=8$, then $9 \mathrm{~d}=72$. The multiples of it and less than 288 are 144 and 216. But these numbers do not have a divisor with the sum of digits 5.

Additional criteria for checking.

"2 or 3 points" Only the correct answer is provided

#
# Task 8.2

For a natural number $N$, all its divisors were listed, and then the sum of digits for each of these divisors was calculated. It turned out that among these sums, all numbers from 1 to 9 were found. Find the smallest value of $\mathrm{N}$.

## Number of points 7