3. Determine the smallest natural number $n$, for which the following holds: If some natural number has at least $n$ three-digit multiples, then 840 is one of them.

(Michal Rolínek)
Solution. From the factorization $840=2^{3} \cdot 3 \cdot 5 \cdot 7$ we see that the smallest natural number that is not a divisor of 840 is the number $9=3^{2}$. The three-digit multiples of the number 9 are

$$
108=9 \cdot 12, 117=9 \cdot 13, \ldots, 999=9 \cdot 111 \text {, }
$$

in total there are $111-12+1=100$ (and the number 840 is certainly not among them). The example of the number 9 thus shows that no number $n \leqq 100$ has the required property.

In the second part of the solution, we will show that the number $n=101$ already has the required property. For this, it is enough to verify that every natural number that has at least 101 three-digit multiples is less than 9 - then it is, as we already know, a divisor of the number 840. In other words, it is enough to verify that every natural number $a \geqq 10$ has at most 100 three-digit multiples. This is easy: the number of all its multiples less than 1000 is at most $999 / a \leqq 99.9100$ evaluate 3 points, of which the observation that the smallest number that is not a divisor of 840 is the number 9 is worth 1 point, and 1 point is given for the correct determination of the number of three-digit multiples of the number 9. However, it is not enough to simply divide $900: 9=100$ without further comment.

The proof that $n=101$ satisfies the condition is also evaluated 3 points. For this part of the solution, give only 1 point if the solver finds that the numbers from 1 to 8 have at least 112 three-digit multiples each, and from this incorrectly concludes that $n=112$ is the desired value.