3. The number n is the product of three different prime numbers. If we increase the two smaller ones by 1 and leave the largest one unchanged, the product of the three will increase by 915. Determine the number n.
SOLUTION. Let $n=p q r, p<q<r$. The equation $(p+1)(q+1) r=p q r+915$ can be equivalently rearranged to the form $(p+q+1) \cdot r=915=3 \cdot 5 \cdot 61$, from which it follows that the prime number $r$ can only take some of the values 3, 5, and 61. For $r=3$, however, from the last equation we get $(p+q+1) \cdot 3=3 \cdot 5 \cdot 61$, or $p+q=304$. This is in contradiction with the fact that $r$ is the largest. Similarly, we find that $r$ cannot be 5 either. Therefore, $r=61$ and $p+q=14$. By trying all possibilities for $p$ and $q$, we get $p=3, q=11, r=61$ and $n=3 \cdot 11 \cdot 61=2013$.

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