Question 25

This question tests two skills.

First, you must know that an even number is divisible by 2. Hence, the last number of the extension must be even (2 only divides evenly into an even number). In this case the extension must end in either 2 or 6.

Second, you must be able to recognize how many different possible orderings there can be for the first three numbers which are:

1, 2, and 3 if the extension is 6; and

1, 3, and 6 if the extension is 2.

Mathematical Solution

Do you remember factorials from high school? (The GMAT is your first real world application of something you learned in high school.) We are determining possible orderings for 3 numbers.

3! = 3 * 2 * 1 = 6. Therefore, there are six possible orderings if the extension ends in 6 and six possible orderings if the extension ends in 2. Since 6 + 6 = 12, the answer is (C).

Multiple Choice Is Your Friend Solution

Let's say the extension ends in 6. The first three numbers are some ordering of 1, 2 and 3. How many such orderings exist?

Let's put 1 in position 1. Our possible orderings are now
1, 2, 3, 6 or 1, 3, 2, 6 (two possible orderings)

Let's put 2 in position 1. Our possible orderings are now 2, 1, 3, 6 or 2, 3, 1, 6. (two possible orderings)

Time out to check the answer choices! We are at 4 possible orderings already. Since the extension could also end in 2 we can always reverse the 2 and 6. This doubles our possible orderings bringing us from 4 to 8. Hence, choices (A) and (B) can be eliminated.

Now, let's put 3 in position 1. Our possible orderings are now
3, 1, 2, 6 or 3, 2, 1, 6 (two possible orderings)

Again though we can reverse the 6 and 2 giving us two more possible orderings.

Our total number of possible orderings is now 8 + 4 = 12.

Hence, the correct answer is (C).