2007 Cyprus MO/Lyceum/Problem 10

Problem

The volume of an orthogonal parallelepiped is $132\;\mathrm{cm}^3$ and its dimensions are integers. The minimum sum of the dimensions is

$\mathrm{(A) \ } 27\;\mathrm{ cm}\qquad \mathrm{(B) \ } 19\;\mathrm{ cm}\qquad \mathrm{(C) \ }20\;\mathrm{ cm}\qquad \mathrm{(D) \ } 18\;\mathrm{ cm}\qquad \mathrm{(E) \ } \mathrm{None\ of\ these}$

Solution

$\displaystyle V = lwh$, so we want to minimize the sum of three integers whose product is $\displaystyle 132 = 2^2\cdot3\cdot11$. To do this, the factors must be as close together as possible. Therefore, none of the factors will be $2$, and one will likely be $11$. This implies that the factors are minimized when they are $3,\ 4,\ 11$, and the answer is $3 + 4 + 11 = 18 \Longrightarrow \mathrm{D}$.

See also

2007 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 9
Followed by
Problem 11
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