Divisors From Prime Factorization Example

Hello everyone. So in this example, we're going to count the number of even divisors of the number $1280$.

Now when you see the word “divisors”, you should be thinking maybe using divisors from prime factorization, so DFPF. But in order to do that, you need to prime factor $1280$. So first, we factor $1280$.

$$1280=4\cdot 320=4\cdot 32\cdot 10=2^2\cdot 2^5\cdot 2\cdot 5=2^8\cdot 5$$

Now DFPF, so Divisors From Prime Factorization, says that the divisors of this number of the form, $2^a5^b$ with $0\leq a\leq 8$, and $0\leq b\leq 1$.

This gives us all the divisors of $1280$. However, to get just the even factors, we need at least one copy of $2$, and hence we will increase the lower bound on $a$ to $1$, and thus $1\leq a \leq 8$.

So what are the possible values now for $a$? Well $a$ is now between $1$ and $8$, and so thus, we have $7+1$ possible factors (the $+1$ is because we count both endpoints, so we take $8-1=7+1=8$ for both endpoints) and it gives us the total number of possibilities for the power of $2$.

And we're going to take $1+1=2$, and that's the total possibilities for the factors of $5$, right? It's either $5^0$, or $5^1$. That's $8\cdot 2=16$ possible even divisors of $1280$.

So you might want to practice writing down the $16$ possibilities just to get some extra divisor from prime factorization help. So thank you very much, and hopefully this gives you an idea of how to use DFPF to compute the divisors of a number.