Prime Factors on the GMAT

Some of the most common Number Properties questions on the GMAT require you to recognize how many times a certain number is a factor of another number.

For example “If 81 =3^x, then x = ?”

Most students answer this question correctly because they recognize that 81 can be re-written as 3⁴, thus making the answer 4.

This may seem like a simple concept, but, as is the case with most GMAT concepts, difficulties arise when the test-makers disguise the concept that a question is testing.

Look at the following question (adapted from a GMATPrep CAT):

If 2^x3^y = 288, what is x + y?

This question stumps a lot of students because it’s equating an exponential product with an integer. However, close scrutiny of the left side of the equation reveals a key fact: We’re dealing with prime numbers! If these two numbers are equal, then, by definition, their prime factorization must be the same. So, if we figure out the prime factorization of 288, we can determine how many times 2 is a factor of 288 and how many times 3 is a factor of 288.

Doing the prime factorization, we arrive at: 288 = 2⁵3². Thus, x = 5, y = 2, and x + y = 7.

Let’s look at another example adapted from a GMATPrep CAT:

If 5^m4¹⁸ = 2(10)³⁵, then m = ?

This question fundamentally asks how many times 5 appears in the prime factorization of 2(10)³⁵. When most of my students see this question, their initial reaction is to break both sides of the equation into their prime factors, thus determining how many times 5 AND 2 appear in the prime factorization of each number. This approach will work, but it requires unnecessary and time-consuming steps that can be avoided. The key is recognizing that the 4¹⁸ on the left side of the equation and the 2 on the right side of the question are irrelevant.

Why? Because we’re concerned with how many times 5 appears in the prime factorization of the product on the right. Since 2 isn’t a factor of 5, determing how many times 2 or 4 appears in the prime factorization won’t tell us anything we care about. What we should focus on is 10³⁵ since that’s the product that will have 5 as a factor. If this number has 10 as a factor 35 times, then the factors of ten (5 and 2) must appear 35 times as well. Thus, there are 35 fives in the prime factorization of this number.

To get more practice on this tricky concept, take a look at Problem Solving Question 107 in the Official Guide, 12^th Edition.