4. The product of three consecutive natural numbers is 65 times greater than their sum. What are these numbers?
4. The product of three consecutive natural numbers is 65 times greater than their sum. What are these numbers?

## Solution.

Let $n, n+1, n+2$ be three consecutive natural numbers.

Then we have:

$$
\begin{gathered}
n \cdot(n+1) \cdot(n+2)=65 \cdot(n+n+1+n+2) \\
n \cdot(n+1) \cdot(n+2)=65 \cdot 3 \cdot(n+1)
\end{gathered}
$$

The number $n+1$ is different from zero, so we can divide the entire equation by $n+1$.

We get

$$
n \cdot(n+2)=195
$$

The number 195 needs to be expressed as the product of two natural numbers that differ by 2.

The prime factorization of 195 is $195=3 \cdot 5 \cdot 13$,

$$
\begin{gathered}
195=13 \cdot 15 \\
n=13 .
\end{gathered}
$$

Note 1: A student can denote three consecutive natural numbers in some other way, for example, $n-1, n$, and $n+1$.

Note 2: If a student determines the three numbers by trial and error according to the given conditions, without arguing that there are no other solutions, they receive a total of 2 POINTS.