$[. A$.  Given three distinct natural numbers, one of which is equal to the half-sum of the other two.  Can the product of these three numbers be a perfect 2008th power of a natural number?}
For example, $6^{669}, 2 \cdot 6^{669}, 3 \cdot 6^{669}$. Their product is $6 \cdot 6^{3 \cdot 669}=6^{2008}$.

## Answer

It can.

## $\begin{array}{ll}\text { Problem } 111651 & \text { Topics: } \\ \text { [Fundamental Theorem of Arithmetic. Factorization into prime factors }]\end{array}$  Difficulty: 3+ $[\quad$ Examples and counterexamples. Constructions

Author: Galperin G.A.

Does there exist an arithmetic progression of five different natural numbers, the product of which is a perfect 2008th power of a natural number?

## Solution

For example, $120^{803}, 2 \cdot 120^{803}, 3 \cdot 120^{803}, 4 \cdot 120^{803}, 5 \cdot 120^{803}$. Their product is $120 \cdot 120^{5 \cdot 803}=120^{4016}=$ $\left(120^{2}\right)^{2008}$.

## Answer

It exists.