# Manhattan GMAT Challenge Problem of the Week – 23 Nov 2010

by on November 23rd, 2010

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## Question

A rectangular solid has side lengths 3, 4, and z, where z is an integer. The total surface area of this solid is the same as that of a cube with side length s, where s is an integer. If z is the lowest possible integer that fits these conditions, what is z?

(A) 3

(B) 5

(C) 6

(D) 7

(E) 9

The surface area of a rectangular solid is the total area of all of its faces. Since the side lengths are 3, 4, and z, there will be two faces with area 12 (= 3×4), two faces with area 3z, and two faces with area 4z. Thus, the total surface area of this solid will be 2 × (12 + 3z + 4z) = 24 + 14z.

Meanwhile, since a cube has 6 faces, its surface area is 6 times the area of one face, or .

We now set these expressions equal to each other.

We need to find two integers, z and s, that fit this equation, and z needs to be its lowest possible value. At this point, we should switch to testing numbers—although we can save ourselves some work if we notice that z must be a multiple of 3. (Why? Notice that the right side, , is a multiple of 3, and on the left side, 12 is a multiple of 3. That means 7z must be a multiple of 3, and since 7 is not a multiple of 3 itself, z must be a multiple of 3.) Also, z must be a positive integer (otherwise, the rectangular solid would not be a solid), so we can start with z = 3 and check.

If , meaning that , but 11 is not a perfect square. No good.
If If , meaning that , but 18 is not a perfect square. No good.
If If , meaning that , so s = 5. This fits.

Since we are asked for s, we must pick 5. Be careful not to pick z (which would be answer E). The correct answer is (B).

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