Difference between revisions of "Mock AIME 1 Pre 2005 Problems/Problem 9"
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== Solution == | == Solution == | ||
| − | <cmath>\begin{align*} \frac {1}{p} + \frac {1}{q} + \frac {1}{r} + \frac {360}{pqr} & = | + | <cmath>\begin{align*} \frac {1}{p} + \frac {1}{q} + \frac {1}{r} + \frac {360}{pqr} & = 1 \\ |
pq + pr + qr + 360 & = pqr \\ | pq + pr + qr + 360 & = pqr \\ | ||
360 & = pqr - pq - pr - qr \\ | 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) \\ | 385 & = (p - 1)(q - 1)(r - 1) \\ | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
| − | 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> | + | |
| + | 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>. | ||
== See also == | == See also == | ||
| − | {{Mock AIME box|year=Pre 2005|n=1|num-b= | + | {{Mock AIME box|year=Pre 2005|n=1|num-b=8|num-a=10|source=14769}} |
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
Latest revision as of 20:47, 21 February 2010
Problem
and 
are three non-zero integers such that 
and
![\[\frac{1}{p} + \frac{1}{q} + \frac{1}{r} + \frac{360}{pqr} = 1\]](http://latex.artofproblemsolving.com/6/8/8/68867ee7b08e34c84bd7054c465e0b101d58f786.png)
Compute 
.
Solution
From here, you can factor 
as 
, giving corresponding values of 
and 
. The answer is 
.
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 | ||
