# ASVAB Preparation: Least Common Multiples

The ASVAB will likely have questions asking about common multiples. A *common multiple* is a number that is a multiple of two or more numbers. For example, 20, 30, and 40, are common multiples of the numbers 5 and 10.

The *least common multiple* (LCM) of two or more numbers is the smallest number (not zero) that’s a multiple of both or all the numbers. The LCM is useful in solving many math problems — especially those involving fractions.

One way to find the LCM is to list the multiples of each number, one at a time, until you find the smallest multiple that’s common to all the numbers.

Find the LCM of 45 and 50.

**Multiples of 45:**45, 90, 135, 180, 225, 270, 315, 360, 405, 450**Multiples of 50:**50, 100, 150, 200, 250, 300, 350, 400, 450

The LCM of 45 and 50 is 450.

That’s rather cumbersome, isn’t it? Wouldn’t it be great if you had an easier way? You do: The easiest way to find the LCM is first to list the prime factors of each number:

The prime factors for 45 are 3 × 3 × 5.

The prime factors for 50 are 2 × 5 × 5.

Then multiply each factor the greatest number of times it occurs in either number. If the same factor occurs more than once in both numbers, you multiply the factor the greatest number of times it occurs.

For example, 5 occurs as a prime factor of both 45 (where it occurs once) and 50 (where it occurs twice); the two occurrences in the factorization of 50 trump the single occurrence in the factorization of 45. The number 3 occurs two times, 5 occurs two times, and 2 occurs once, so you have 3 × 3 × 5 × 5 × 2 = 450.

Checking your answer to see whether the original numbers divide evenly into the LCM you calculate is always a great idea. You can in fact divide 45 and 50 evenly into 450, so you’re good to go in this example.

Now that you’re getting the hang of it, try another one:

What is the least common multiple of 5, 27, and 30?

List the prime factors of each number:

**Prime factors of 5:**5**Prime factors of 27:**3 × 3 × 3**Prime factors of 30:**2 × 3 × 5

The number 3 occurs a maximum of three times, 5 occurs a maximum of one time, and 2 occurs a maximum of one time: 3 × 3 × 3 × 5 × 2 = 270. Check your answer by seeing whether 5, 27, and 30 can all divide evenly into 270.