1. For any three-digit number, we determine its remainders when divided by the numbers 2, 3, 4, ..., 10 and then sum the nine resulting numbers. Find the smallest possible value of such a sum.
SOLUTION. Let $S(n)$ denote the sum of the mentioned remainders of the three-digit number $n$. We will explain why $S(n) \geqq 3$.

- For odd $n$, $S(n) \geqq 5$ (consider the remainders when dividing by even numbers $2,4,6,8$, 10). Hence, let $n$ be even from now on.
- If $4 \nmid n$, then $S(n) \geqq 4$ ( $n$ gives a remainder of at least 2 when divided by 4 and 8). Let $n$ be divisible by 4 from now on.
- If $8 \nmid n$, then $S(n) \geqq 4$ (remainder 4 when divided by 8). Therefore, let $n$ be divisible by 8 from now on.
- If $3 \nmid n$, then $S(n) \geqq 3$ ( $n$ gives a remainder of at least 1 when divided by $3,6,9$). Let $n$ be divisible by 8 and 3 from now on.
- If $9 \nmid n$, then $S(n) \geqq 3$ (remainder of at least 3 when divided by 9). Let $8 \mid n$ and $9 \mid n$ from now on.
- If $5 \nmid n$, then $S(n) \geqq 3$ (remainder of at least 1 when divided by 5 and at least 2 when divided by 10).

Therefore, assume that $5|n$, $8| n$, and $9 \mid n$. Then only the numbers 360 and 720 come into consideration, for which $S(360)=3$ and $S(720)=9$. This proves the inequality $S(n) \geqq 3$. At the same time, we have found that $S(n)=3$ for example for $n=360$. (It is also $S(840)=3$.)