Manhattan GMAT Challenge Problem of the Week – 23 Nov 2010

by on November 23rd, 2010

Here is a new Challenge Problem! If you want to win prizes, try entering our Challenge Problem Showdown. The more people enter our challenge, the better the prizes.


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 6s^2.

We now set these expressions equal to each other.

24 + 14z = 6s^2
12 + 7z = 3s^2

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, 3s^2 , 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 z = 3, 12 + 7z = 12 + 21 = 33 = 3s^2, meaning that s^2 = 11, but 11 is not a perfect square. No good.
If If z = 6, 12 + 7z = 12 + 42 = 54 = 3s^2, meaning that s^2 = 18, but 18 is not a perfect square. No good.
If If z = 9, 12 + 7z = 12 + 63 = 75 = 3^s, meaning that s^2 = 25, 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).

Special Announcement: If you want to win prizes for answering our Challenge Problems, try entering our Challenge Problem Showdown. Each week, we draw a winner from all the correct answers. The winner receives a number of our our Strategy Guides. The more people enter, the better the prize. Provided the winner gives consent, we will post his or her name on our Facebook page.


  • :) thanks

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

Ask a Question or Leave a Reply

The author Caitlin Clay gets email notifications for all questions or replies to this post.

Some HTML allowed. Keep your comments above the belt or risk having them deleted. Signup for a Gravatar to have your pictures show up by your comment.