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

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## Question

A rectangular solid has side lengths 3, 4, and

z, wherezis an integer. The total surface area of this solid is the same as that of a cube with side lengths, wheresis an integer. Ifzis the lowest possible integer that fits these conditions, what isz?

(A) 3

(B) 5

(C) 6

(D) 7

(E) 9

## Answer

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 3*z*, and two faces with area 4*z*. Thus, the total surface area of this solid will be 2 × (12 + 3*z* + 4*z*) = 24 + 14*z*.

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 7*z* 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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## 2 comments

domenico on November 23rd, 2010 at 1:16 pm

thanks

Roy on September 29th, 2011 at 2:01 pm

The questions asks for Z, NOT S. The correct answer is E.