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1998 AIME Problems/Problem 14

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Problem

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

Solution

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

Let’s solve for :

(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 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 (ProblemsResources)
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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