Difference between revisions of "Mock AIME 1 Pre 2005 Problems/Problem 9"

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385 & =  (p - 1)(q - 1)(r - 1) \\
 
385 & =  (p - 1)(q - 1)(r - 1) \\
 
\end{align*}</cmath>
 
\end{align*}</cmath>
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From here, you can factor <math>385</math> as <math>5 \cdot 7 \cdot 11</math>, giving corresponding values of <math>6, 8,</math> and <math>12</math>. The answer is <math>6 \cdot 8 \cdot 12=576</math>.
 
From here, you can factor <math>385</math> as <math>5 \cdot 7 \cdot 11</math>, giving corresponding values of <math>6, 8,</math> and <math>12</math>. The answer is <math>6 \cdot 8 \cdot 12=576</math>.
  

Latest revision as of 21:47, 21 February 2010

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 \\  & =  (p - 1)(q - 1)(r - 1) - (p + q + r) + 1 \\  & =  (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 $12$. The answer is $6 \cdot 8 \cdot 12=576$.

See also

Mock AIME 1 Pre 2005 (Problems, Source)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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