Prime Factorization is a fundamental concept of Number Theory, one of the main topics that are covered on MOEMS tests. Let’s have a look at what it is and the various ways it is used.

• Find the prime factorization of 12. How many factors does 12 have?

Prime factorization involves splitting a number into its factors that are prime. To do this, we can divide the original number by a prime number and repeat until we end up with a prime number. Let’s do this with 12.

First, we can divide 12 by 3 and get 4. We can divide 4 by 2 and get 2. So, the prime factorization of 12 is 2 x 2 x 3, or 22 x 3.

While prime factorization may seem like it’s useful, it is arguably the most important tool in competition math.

Now, it is time to find the number of factors of 12. A trick to this is recognizing that every factor has to either be 1, a prime factor, or a product of prime factors.

Each factor can have zero, one, or two 2s. Additionally, each factor can have zero or one 3. Since there are three options for selecting the amount of 2s and two options for selecting the number of threes, there are a total of 3 x 2 = 6 factors of 12.

We can derive a simple rule from this. The number of factors of a number is equal to the product of one plus each of its prime factors’ exponents. Here it would be (2 + 1) x (1 + 1) = 6.

Let’s try a more intuitive problem:

• A college class has 120 students. The professor is assigning a group project. How many different ways are there to select a number of group members per group such that there are no people remaining after all groups have been chosen? The whole class cannot be a group and students cannot work individually.

This problem seems complicated, but the wording is hiding the underlying simplicity of the problem. Since we want the number of possible group sizes without remainders, any possible group size must be a factor of 120 so that there will be no students remaining.

So, we need to find the prime factorization of 120 and get the number of factors. Then, we need to subtract 2 for the two group sizes disallowed in the problem.

The prime factorization of 120 is 23 x 3 x 5. So, the total number of factors is (3 + 1) x (1 + 1) x (1 + 1) = 4 x 2 x 2 = 16. Subtracting two, we get 14.

As you can see, prime factorization is often disguised. When in doubt, take the prime factorization and work from there.