Complicated Factoring Problems

[Chud]

I have been getting tripped up on complex factoring problems. Simple questions but usually complex answers. Can anybody explain?

How many factors does 36^2 have?

a) 2
b) 8
c) 24
d) 25
e) 26

If the answer provided was in greek, it would read the same to me. Please help...Thanks!

[Guest]

How many factors does 36^2 have?

First, find the prime factors: 36^2 = (6*6)^2 = (2^2*3^2)^2 = 2^4 *3^4

This means there are 5 possibilities for both factors (between 0 and 4 of each) and 5*5 = 25. This captures the case of 2^0*3^0 =1

Not that you would list these on the exam, but here are all the factors: 1,2,4,8,16,3,9,27,81,6,12,24,48,18,36,72,144,54,108,216,432,162,324,648,1296

[StaceyKoprince]

Please cite the source of the problem - if you do not do so, we will have to delete it. Thanks!

[Chud]

The Problem is from one of the MGMAT practice tests.

Can you further go into why it is 5 x 5 ?

[Guest]

You can also approach the question like this:

36^2 = (6^2)^2 = 6^4 = (2*3)^4 = 2^4 * 3^4
No. of factors = (4+1)(4+1) = 25

[StaceyKoprince]

As an earlier poster stated, first break 36^2 down into its prime factors. I recommend writing out the below on paper so you can follow each step - understand each one before you go onto the next.
36^2 => break down 36
(6*6)^2 => distribute
6^2 * 6^2 => break down each 6
(2*3)^2 * (2*3)^2 => distribute
2^2 * 3^2 * 2^2 * 3^2 => combine the base-2's and base-3's
2^4 * 3^4
So I have four 2's and four 3's.

(Note that you could do this in slightly different orders / arrangements, as other posters have done above, but you'd still end up with four 2's and four 3's.)

These are the prime factors, which make up all of the factors. For example, I can have:
2
2*2
2*2*2
2*2*2*2
etc.

each unique arrangement of these 8 prime factors will give me a different number and each number will be a factor. And, of course, I don't want to forget that 1 will be a factor, too, even though it's not represented in these 8 prime factors. 1 is the case of using no 2's or 3's to create the factor.

If I want to know how many unique factors I could get, I could write them out (and, if you had to, that's what you'd do - but don't actually do the multiplication, just write them out as I did just above). Or I could just know the rule for calculating this given that I have four 2's and four 3's.

I take my first unique prime factor, 2, and say that any factor I create could use up to four 2's, including the possibility that I don't use any 2's at all. So that's 5 possibilities for how I can use my 2's to create a factor: no 2's, one 2, two 2's, three 2's, or four 2's.

The same thing is true of my other unique prime factor, 3. I have up to four of them to use, or I might not use any of them, so I have 5 ways I can use my 3's.

Once I figure that out, the rule is simply to multiply those two possibilities: 5*5 = 25. This will cover all of the unique factors I can create while not calculating overlapping factors (eg, 2*3 is the same factor as 3*2, 6, so I don't want to count that twice).