4. For any natural number K, we determine its remainders when divided by each of the ten natural numbers $2,3,4, \ldots, 11$ and sum these ten remainders (some of which may be zero). Determine all such numbers less than 25000 that have the smallest possible sum of these remainders. (We do not consider zero as a natural number.)

The regional round of category B takes place

## on Tuesday, April 7, 2009

so that it starts in the morning and contestants have 4 hours of pure time to solve the problems. For each problem, a contestant can earn 6 points, and a successful solver is a student who scores 10 points or more. Allowed aids are writing and drawing supplies, school MF tables, and calculators without a graphical display. These details will be communicated to the students before the start of the competition.
4. Consider the natural number $n25000$.

We will show that the smallest possible sum is 1, and at the same time find all numbers $n$ less than 25000 with this property.

If the relevant sum is 1, all remainders $r_{k}$ except one are equal to 0, and there is thus exactly one $d \in \mathrm{M}$ such that $r_{d}=1$. We will show that $d=7$ or $d=11$. It is clearly impossible for $d \leqq 5$, as a non-zero remainder would correspond to the number $2d \in \mathrm{M}$. If the remainder 1 corresponds to one of the numbers $d=6,8,9,10$, it would necessarily correspond to one of the numbers 2 or 3.

If $d=7$, the sought number must be a multiple of all numbers in $\mathrm{M} \backslash\{7\}$, i.e., a multiple of the number 3960. This number gives a remainder of 5 when divided by 7, and the remainder 1 is given by its triple $n=3 \cdot 3960=11880$, which satisfies the conditions of the problem, and generally by any $(3+7a)$-multiple; however, the next multiple $10 \cdot 3960$ with the appropriate remainder is already greater than 25000.

If $d=11$, the sought number must be a multiple of all numbers in $\mathrm{M} \backslash\{11\}$, i.e., a multiple of the number 2520. Since this number gives a remainder of 1 when divided by 11, the only number that satisfies the condition for $d=11$ is 2520 (the next multiple $(1+11) \cdot 2520$ with the appropriate remainder is already greater than 25000).

Conclusion: The sought numbers are two, namely 11880 and 2520.

For a complete solution, award 6 points, for proving that the sum of remainders is at least one, award 1 point, for proving that the sought number is divisible by all numbers in $\mathrm{M}$ except 7 or 11, for which it gives a remainder of 1, award 3 points, for finding the corresponding numbers 11880 and 2520, award 1 point each. If the solver shows that the sum of remainders is at most 1, and by providing one of the suitable numbers shows that the smallest sum of remainders is indeed 1, evaluate their solution as 3 points (since the task was to find all suitable numbers).