Difference between revisions of "1998 AIME Problems/Problem 14"
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== Problem == | == Problem == | ||
+ | An <math>m\times n\times p</math> rectangular box has half the volume of an <math>\displaystyle (m + 2)\times(n + 2)\times(p + 2)</math> rectangular box, where <math>m, n,</math> and <math>p</math> are integers, and <math>m\le n\le p.</math> What is the largest possible value of <math>p</math>? | ||
== Solution == | == Solution == | ||
+ | <div style="text-align:center;"><math>\displaystyle 2mnp = (m+2)(n+2)(p+2)</math></div> | ||
+ | |||
+ | Let’s solve for <math>p</math>: | ||
+ | |||
+ | <div style="text-align:center;"><math>(2mn)p = p(m+2)(n+2) + 2(m+2)(n+2)</math><br /> | ||
+ | <math>[2mn - (m+2)(n+2)]p = 2(m+2)(n+2)</math><br /> | ||
+ | <math>p = \frac{2(m+2)(n+2)}{mn - 2n - 2m - 4}</math></div> | ||
+ | |||
+ | For the denominator, we will use a factoring trick (colloquially known as [[Simon's Favorite Factoring Trick|SFFT]]), which states that <math>xy + ax + ay + a^2 = (x+a)(y+a)</math>. | ||
+ | |||
+ | <div style="text-align:center;"><math>p = \frac{2(m+2)(n+2)}{(m-2)(n-2) - 8}</math></div> | ||
+ | |||
+ | Clearly, we want to minimize the denominator, so <math>\displaystyle (m-2)(n-2) - 8 = 1 \Longrightarrow (m-2)(n-2) = 9</math>. The possible pairs of factors of <math>9</math> are <math>\displaystyle (1,9)(3,3)</math>. These give <math>m = 3, n = 11 \displaystyle</math> and <math>m = 5, n = 5</math> respectively. Substituting into the numerator, we see that the first pair gives <math>130</math>, while the second pair gives <math>98</math>. We can quickly test for the denominator assuming other values to convince ourselves that <math>1</math> is the best possible value for the denominator. Hence, the solution is <math>p = 130</math>. | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=1998|num-b=13|num-a=15}} | |
+ | |||
+ | [[Category:Intermediate Algebra Problems]] |
Revision as of 11:21, 9 September 2007
Problem
An rectangular box has half the volume of an
rectangular box, where
and
are integers, and
What is the largest possible value of
?
Solution
![$\displaystyle 2mnp = (m+2)(n+2)(p+2)$](http://latex.artofproblemsolving.com/a/3/4/a3463af31c4d406d36f6d70199b204460a045d7d.png)
Let’s solve for :
![$(2mn)p = p(m+2)(n+2) + 2(m+2)(n+2)$](http://latex.artofproblemsolving.com/d/5/f/d5ff63a88caf1ac4552d9215366e18fab5e144d8.png)
![$p = \frac{2(m+2)(n+2)}{mn - 2n - 2m - 4}$](http://latex.artofproblemsolving.com/f/1/0/f108f6bdc2234b440849dafc9bf269e8e77d0e8d.png)
For the denominator, we will use a factoring trick (colloquially known as SFFT), which states that .
![$p = \frac{2(m+2)(n+2)}{(m-2)(n-2) - 8}$](http://latex.artofproblemsolving.com/b/8/4/b84a61a6b6fdd06d3924cbb85b37bbabef83703f.png)
Clearly, we want to minimize the denominator, so . The possible pairs of factors of
are
. These give
and
respectively. Substituting into the numerator, we see that the first pair gives
, while the second pair gives
. We can quickly test for the denominator assuming other values to convince ourselves that
is the best possible value for the denominator. Hence, the solution is
.
See also
1998 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |