Mock AIME 1 Pre 2005 Problems/Problem 9

Problem

$p, q,$ and $r$ are three non-zero integers such that $p + q + r = 26$ and $\frac{1}{p} + \frac{1}{q} + \frac{1}{r} + \frac{360}{pqr} = 1$ Compute $pqr$ .

Solution

$\begin{align*} \frac {1}{p} + \frac {1}{q} + \frac {1}{r} + \frac {360}{pqr} & = & 1 \\ pq + pr + qr + 360 & = pqr \\ 360 & = pqr - pq - pr - qr \\ 360 & = (p - 1)(q - 1)(r - 1) - (p + q + r) + 1 \\ 360 & = (p - 1)(q - 1)(r - 1) - 25 \\ 385 & = (p - 1)(q - 1)(r - 1) \\ \end{align*}$ From here, you can factor $385$ as $5 \cdot 7 \cdot 11$ , giving corresponding values of $6, 8,$ and $10$ . The answer is $6 \cdot 8 \cdot 10=480$ .[/quote]

Mock AIME 1 Pre 2005 (Problems, Source)
Preceded by Problem 7	Followed by Problem 9
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Mock AIME 1 Pre 2005 Problems/Problem 9

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