# Problem of the Day

If a positive integer is equal to the following product: , where and are distinct prime numbers greater than , how many distinct even factors does the integer have?

A. 1

B. 5

C. 50

D. 100

E. 120

### Solution

[latexpage]

This solution assumes that you are familiar with the basics of counting and prime factorization.

To calculate the number of factors that an integer has, first look at its prime factorization. Â For example, if we want to know how many factors 12 has, we would break it down as $2^2 \cdot 3$. Â If an integer is a factor of 12, it has to be built from the prime factors of 12. Â A complete list of the factors of 12 is 1, 2, 3, 4, 6, 12.

$1=2^0 \cdot 3^0$

$2=2^1 \cdot 3^0$

$3=2^0 \cdot 3^1$

$4=2^2 \cdot 3^0$

$6=2^1 \cdot 3^1$

$12=2^2 \cdot 3^1$

Notice that each factor of 12 is of the form $2^x \cdot 3^y$ where $x$ is either 0, 1, or 2 and y is either 0 or 1. Â So there are three choices for the exponent for 2 and two choices for the exponent of 3. Â Therefore there are $3 \cdot 2$ or $6$ factors of 12. Â Another example: If a number has a prime factorization of $2^4 \cdot 3^5 \cdot 5^2$ then it would have $5 \cdot 6 \cdot 3$ or 90 factors. Â In sum, to find the number of factors a number has, just add 1 to each exponent in its prime factorization and then multiply all of the resulting numbers together.

In this problem we’re asked to find the number of distinct *even*Â factors. Â An even number must contain at least one factor of two. Â Any factor of this number must be in the form $2^xb^yc^z$ where $x$ is any integer from 1 to 5, inclusive, $y$ is any integer from 0 to 3, and $z$ is any integer from 0 to 4. Â Again, $x$ cannot be zero because then it would not be an even factor. Â So, there are $5 \cdot 4 \cdot 5$ or 100 even factors of this number.

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