1998 AIME Problems/Problem 14

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Problem

An $m\times n\times p$ rectangular box has half the volume of an $\displaystyle (m + 2)\times(n + 2)\times(p + 2)$ rectangular box, where $m, n,$ and $p$ are integers, and $m\le n\le p.$ What is the largest possible value of $p$ ?

Solution

$\displaystyle 2mnp = (m+2)(n+2)(p+2)$

Let’s solve for $p$ :

$(2mn)p = p(m+2)(n+2) + 2(m+2)(n+2)$

$[2mn - (m+2)(n+2)]p = 2(m+2)(n+2)$

$p = \frac{2(m+2)(n+2)}{mn - 2n - 2m - 4}$

For the denominator, we will use a factoring trick (colloquially known as SFFT), which states that $xy + ax + ay + a^2 = (x+a)(y+a)$ .

$p = \frac{2(m+2)(n+2)}{(m-2)(n-2) - 8}$

Clearly, we want to minimize the denominator, so $\displaystyle (m-2)(n-2) - 8 = 1 \Longrightarrow (m-2)(n-2) = 9$ . The possible pairs of factors of $9$ are $\displaystyle (1,9)(3,3)$ . These give $m = 3, n = 11 \displaystyle$ and $m = 5, n = 5$ respectively. Substituting into the numerator, we see that the first pair gives $130$ , while the second pair gives $98$ . We can quickly test for the denominator assuming other values to convince ourselves that $1$ is the best possible value for the denominator. Hence, the solution is $130$ .

1998 AIME (Problems • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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1998 AIME Problems/Problem 14

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Problem

Solution

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