The sum of two numbers is 2160; their least common multiple $k$ is 9828. Which are these numbers?
Let the sought numbers be $A$ and $B$, their greatest common divisor $D$, thus

$$
A=a D \text { and } B=b D \text { and their least common multiple is } a b D
$$

Here, $a$ and $b$ are already coprime numbers. According to our data,

$$
A+B=(a+b) D=2160 \quad \text { and } \quad a b D=9828
$$

Since $a$ and $b$ are coprime numbers, it is necessary that $a+b$ and $a b$ are also coprime. If $a+b$ and $a b$ had a common divisor $p$, then $p$ would be a divisor of one of the factors of $a b$, say $a$; since $p$ is also a divisor of $a+b$, it must also be a divisor of $b$, meaning: $p$ would be a common divisor of both $a$ and $b$. This contradicts our assumption that $a$ and $b$ are coprime.

From this, it follows that $D$, i.e., the greatest common divisor of $A$ and $B$, must also be the greatest common divisor of 2160 and 9828. However,

$$
2160=2^{4} \cdot 3^{3} \cdot 5 \quad \text { and } \quad 9828=2^{2} \cdot 3^{3} \cdot 7 \cdot 13 \text {, and thus } \quad D=2^{3} \cdot 3^{3}=108
$$

Accordingly,

$$
a+b=20 \quad \text { and } \quad a b=91
$$

Thus, $a$ and $b$ are the roots of the equation

$$
x^{2}-20 x+91=0
$$

$$
a=7 \quad \text { and } \quad b=13 \quad \text { or } \quad a=13 \quad \text { and } \quad b=7 \text {. }
$$

Essentially, we get one set of numbers that meets the requirements of the problem:

$$
7 \cdot 108=756 \text { and } 13 \cdot 108=1404
$$

Löke Endre (Premontrei g. VI. o. Keszthely)

Note. $a b=91$ can be interpreted as 91 being factored into two factors; these factors should be coprime and their sum should be 20. The factors of 91 are: $1,7,13,91$.

Among these, 7 and 13 satisfy the requirements.