8.5. For a natural number $N$, all its divisors were listed, 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 $N$.
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 $N$ must have a divisor with the sum of digits 9, $N$ is divisible by 9. Now consider the divisor $d$ with the sum of digits 8. $d$ is not divisible by 3, so the numbers $d$ and 9 are coprime, which means $N$ is divisible by $9d$. If $d \geq 32$, then $9d \geq 288$, that is,
$N \geq 288$. Therefore, it remains to check $d=26, d=17$, and $d=8$.

If $d=26$, then $9d=234$. This number does not have a divisor with the sum of digits 5, and any number divisible by it is greater than 288.

If $d=17$, then $9d=153$. This number does not have a divisor with the sum of digits 2, and any number divisible by it is greater than 288.

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

Grading criteria. “+” A complete and well-reasoned solution is provided “士” A generally correct reasoning is provided, containing minor gaps or inaccuracies

“Ғ” Only the correct answer is provided

“-” The problem is not solved or solved incorrectly