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.
Well so here we have 1280,
now I notice immediately that 4 is a factor.
Right? So 4 goes into 12 three times.
No remainder. Then 4 goes into 8 twice,
and 4 goes into 0 zero times.
So if we factor out the 4 we have 4 times 320.
320 is easy to see, that's 32 times 10.
and all of these numbers are small enough that I can prime factor them.
2 squared, times 2 to the 5, times 2, times 5,
that's 2 to the 8 times 5.
Now DFPF, so divisors from prime factorization,
says that the divisors of this number of the form,
“2 to the a times 5 to the b”
with a being a number between 0 and 8,
and b being a number between 0 and 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 is less than or equal to a which is less than or equal to 8.
So, what are the possible values now for a?
Well a is now between 1 and 8, and so thus, we have 7 plus 1 possible factors
(the plus 1 is because we count both endpoints, so
we take 8 minus 1 that's 7 plus 1 for both endpoints),
and that gives us the total number of possibilities for the power of 2.
And we're going to take 1 plus 1 which is equal to 2,
and that's the total possibilities for the factors of 5, right?
It's either 5 to the 0, or 5 to the 1.
It's 1 plus 1, again the difference plus 1 for the fact that we count both endpoints.
That's 8 times 2 and that equals 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.