Are There Better Ways To Find GCF and LCM?

Some of you may have been exposed to this in school. Some of you, like me, were not directly taught this method. Just a heads up.

When I was in fourth or fifth grade, I learned the terms greatest common factor and least common multiple.

For those unfamiliar, the greatest common factor (GCF) refers to the largest number that will cleanly divide (leave no remainder) two or more specific numbers. For example, the biggest number that goes into 2 and 3 is 1, so 1 is the GCF of 2 and 3.

The least common multiple (LCM) refers to the smallest number that can be cleanly divided (leave no remainder) by two or more specific numbers. For example, the smallest number divisible by 2 and 3 is 6, so 6 is the LCM of 2 and 3.

The way I learned to find the GCF of two numbers is as follows.

First, take two numbers. Our numbers will be 8 and 10.

Next, list the factors of each number. For 8, we have factors 1 and 8 (1*8=8) and 2 and 4 (2*4=8). For 10, we have factors 1 and 10 (1*10=10) and 2 and 5 (2*5=10). For each number, we write the factors in increasing order.

Now we circle the factors that 8 and 10 have in common. We notice that there are two factors, 1 and 2, which are factors of both 8 and 10.

But only one can be the GCF. As you may have guessed, we want the greatest number, and that happens to be 2. This means that the GCF of 8 and 10 is 2.

What about the LCM? We start again by writing our numbers 8 and 10.

Since we're considering multiples instead of factors, we want to list the multiples of each number. For 8, we get 1*8=8, 2*8=16, 3*8=24, 4*8=32, 5*8=40, 6*8=48 and so on. For 10, we get 1*10=10, 2*10=20, 3*10=30, 4*10=40, 5*10=50 and so on.

Now if we look through the multiples, we find that they have 40 in common. In reality, the two numbers have infinitely many multiples in common, but 40 is the lowest in value, making it the least common multiple.

Those ways are effective, but once you get to large numbers, they become terribly inefficient. Is there a better way? Let's start with 8 and 10 and figure out the GCF a new way.

We need to factor each number into its primes. We can make 8 into 2*4, and 2 is prime. We can make 4 into 2*2, both of which are prime. So 8 is made of 2*2*2 or 2^3.

We can factor 10 into 2*5, both of which are prime.

Now we need to compare quantities of numbers. For 8 (2^3), we have three 2s. For 10 (2*5), we have one 2 and one 5.

What do both have in common? Each has a single 2. Even though 8 has two extra 2s and 10 has one extra 5, we don't include those. We only care about what they have in common.

Let's repeat for LCM. First, prime factorization.

Now we count quantities.

Now we have to compare those quantities.

We include one 2 (because both have one two). We include two extra 2s (because only the 8 had two extra 2s). We include one extra 5 (because only the 10 had one extra 5). This gives 2*4*5, or 40. Notice that if we had included the 2 they had in common twice (once for each), we'd have 80, or 8*10. Sometimes we can just multiply the two numbers and get the LCM, but that is only true if the two numbers are co-prime (sharing no factors).

Let's try two bigger numbers, say 48 and 64. As always, we complete a prime factorization.

Now we compare the quantities and take only what they have in common (four 2s). This gives us a GCF of 16.

We repeat for LCM.

Again, by comparing quantities, we get four 2s in common, one 3 specific to the first and two 2s specific to the second.

Multiplying these out, we get an LCM of 192.

These methods are perhaps most useful in algebra. When adding fractions, like 1/2 and 1/3, you must first find a common denominator (the least common denominator). In that case, we would get 6 to add 3/6 and 2/6. If you're looking at 1/x and 1/(x+1), the problem still remains. Let's look at two factored expressions below.

Now, like always, we count the factors.

Now we take what they have in common, what is specific to the first and what is specific to the second.

Multiplying them out, we get the factored expression below.

To finish the fraction addition, we would simply have to multiply the top of each fraction by what its denominator was missing from the least common denominator. For example, the first expression did not have an (x+3), so the numerator and denominator would be multiplied by (x+3). For 1/2, what is the bottom missing to get to 6? Since it is missing a 3, we multiply 1*3 and 2*3 to get 3 and 6 to give us 3/6, just as we would expect.

Anyway, I hope you got something out of this. If not, well, there's always next time.

If you have any questions, or if anything isn't clear, just let me know in the comments.

I don't exactly recall learning this explicitly. I saw the types of problems in Algebra I and Algebra II to an extent, but I more figured out the method than I was taught it.

The images used here were created by me using Noteshelf for iPad.