1. An unknown number is divisible by exactly four numbers $z$ from the set $\{6,15,20,21,70\}$. Determine which ones.

(Michal Rolínek)
SOLUTION. If the sought four numbers included both 20 and 21 simultaneously, the unknown number would be a multiple of three, four, five, and seven, and thus would also be divisible by $6, 15, 70$, which contradicts the requirements of the problem. Therefore, one of the numbers 20, 21 does not belong to the sought quartet, and thus all three numbers 6, 15, 70 belong to it. Every common multiple of the numbers 6 and 70 is divisible by both three and seven, and thus also by twenty-one. Therefore, the fourth sought number is the number 21.

The four sought numbers are $6, 15, 21, 70$. Their least common multiple is 210, which is not divisible by the remaining number 20. (Thus, the unknown number can also be any multiple of $210 l$, where $l$ is odd.) This solves the problem.

Note. The problem can be solved systematically by first listing all five four-element subsets, i.e., the sets

$$
\begin{gathered}
A=\{15,20,21,70\}, \quad B=\{6,20,21,70\}, \quad C=\{6,15,21,70\}, \\
D=\{6,15,20,70\}, \quad E=\{6,15,20,21\},
\end{gathered}
$$

and determining which of them can satisfy the conditions of the problem, i.e., that there exists a number $N$ whose divisors include all four elements of the considered set, while the fifth omitted number is not its divisor. Thus, we can successively eliminate all sets except $C$, for example, set A: if $N$ is divisible by both numbers $15=3 \cdot 5$ and $20=2 \cdot 10$, it is also divisible by $2 \cdot 3=6$, i.e., $6 \in \mathrm{A}$, which is a contradiction. Similarly, we arrive at contradictions $15 \in \mathrm{B}, 21 \in \mathrm{D}$, and $70 \in \mathrm{E}$.

GUIDING AND SUPPLEMENTARY PROBLEMS:

N1. Determine five natural numbers whose sum is 20 and whose product is 420. [Since $420=2^{2} \cdot 3 \cdot 5 \cdot 7, 1+4+3+5+7=20$ and the sought numbers, which are multiples of 5 and 7, must be single-digit (the sum of the other three numbers is at least 6), two of the sought numbers are directly the numbers 5 and 7, and the remaining three (not necessarily distinct) lie in the set $\{1,2,3,4,6\}$, with their sum being 8 and their product 12, so they are the triplet $1,3,4$.]

N2. A certain natural number has exactly four divisors, whose sum is 176. Determine this number, knowing that the sum of all its digits is 12. [The sought number $n$ is divisible by three and greater than 9, so its four divisors are precisely the numbers $1<3<n / 3<n$. From this, we get $n=129$.]

D1. A given integer is divisible by at least four numbers from the set $\{2,3,5,6,10,15\}$. Prove that it is divisible by each of them. [Notice that $6=2 \cdot 3, 10=2 \cdot 5$, and $15=3 \cdot 5$. Therefore, if the given number is divisible by all three numbers 2, 3, and 5, it is divisible by all the remaining numbers 6, 10, and 15. In the opposite case, it would be divisible by at most two of the numbers 2, 3, and 5, and thus by at least two of the numbers 6, 10, and 15, and thus by each of the prime numbers 2, 3, and 5, which is a contradiction.]