Kaplan GMAT Sample Problem: Prime Factors

By - Aug 12, 06:00 AM Comments [0]

Dealing with prime numbers and prime factors is an essential GMAT skill. Technically it is a skill we learned as early as elementary school/early school age, however just because it was learned in childhood does not mean that these questions are necessarily easy for GMAT test-takers. First of all, dealing with prime factors may not be something we do on a daily basis, so you’ll want to be sure you practice during your studies. Secondly, sometimes the wording or steps along the way can be challenging on the GMAT, not to mention the time limit. Here is a fairly straightforward question, though it may take some time to work through.

Problem:
What is the smallest positive integer that is a multiple of 18, 20, 24, 25 and 30?
(A) 360
(B) 900
(C) 1,800
(D) 2,400
(E) 3,600

Solution:
While taking each answer choice, starting with the smallest, and dividing by each of the five numbers listed until you find one that divides evenly by all of them will get you to the right answer, you can avoid this time consuming process by employing prime factorization.

First, find the prime factors of the five numbers included in the question. These are as follows:
18 = 3 x 3 x 2
20 = 2 x 2 x 5
24 = 2 x 2 x 2 x 3
25 = 5 x 5
30 = 2 x 3 x 5

Second, we need to put these prime factors together in order to find our answer. We want to find the smallest number that is a multiple of all of our given numbers, so let’s start with the smallest one. The smallest multiple of 18 is 18, which equals 3 x 3 x 2. Next, our number must be a multiple of 20, which equals 2 x 2 x 5. Thus, we need 3 x 3 x 2 x 2 x 5 for our number to be a multiple of both 18 and 20, as this number includes all of the prime factors of both 18 and 20. Our third number is 24, which equals 2 x 2 x 2 x 3. However, the number we are building already includes two 2’s and a 3. Therefore, we only need to include one more 2 in order to make our number a multiple of 24. Now we are at 3 x 3 x 2 x 2 x 5 x 2. For our number to be a multiple of 25, we must add in another 5, giving us 3 x 3 x 2 x 2 x 5 x 2 x 5. Finally, to be divisible by 30 our number must include 2 x 3 x 5. However, as 2 x 3 x 5 is can already be found, we do not need to add in any additional numbers.

Lastly, to reach the answer, we calculate 3 x 3 x 2 x 5 x 2 x 5 in the following manner:
3 x 3 x 2 x 2 x 5 x 2 x 5 =
18 x 10 x 10 =
18 x 100 =
1800, which is answer choice (C).

~Bret Ruber

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