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 10:21, 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
Let’s solve for :
![$[2mn - (m+2)(n+2)]p = 2(m+2)(n+2)$](http://latex.artofproblemsolving.com/e/2/c/e2c32ed27109f6e9f84372c378e822f0bf7dc712.png)
For the denominator, we will use a factoring trick (colloquially known as SFFT), which states that 
.
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 | ||
