A10. Three positive integers add to 93 and have a product of 3375 . The integers are in the ratio $1: k: k^{2}$.
What are the three integers?
Solution
$$
3,15,75
$$

Let the three positive integers be $A, A k$ and $A k^{2}$, where $A$ is a positive constant. Therefore $A+A k+A k^{2}=A\left(1+k+k^{2}\right)=93=3 \times 31$. If $k=1$ and $A=31, A^{3} k^{3}=31^{3} \times 1^{3} \neq 3375$. Therefore, since 3 and 31 are prime, $A=3$ and $1+k+k^{2}=31$. Hence, $k=5$ and the three integers are 3,15 and 75 .