Stumped on this quant question, what am I doing wrong?

GRE Quant Problem Solving
1

Hi Greg,

I’d like some clarification on this question:

A positive integer with three distinct prime divisors cannot have how many positive and negative factors?

20
24
36
48
54

Here’s how I approached it:

  1. I picked 3 distinct prime numbers: 2, 3, and 5.

  2. For the first round, I assumed they all had an exponent of 1. The total # of factors, including negatives, is 16.

  3. For the second round, I assumed the 2 has an exponent of 2 while the 3 and 5 has an exponent of 1. This brings the total # of factors to 322*(2 b/c of negative factors) = 24.

  4. This means that option A is likely the answer. But I keep checking.

  5. For the 3rd round, I raised the 2 to a power of 3 and kept the exponents of 3 and 5 as 1. Total factors= 422*2= 32 total factors.

  6. I’m attempting to bring the total # of factors to 36, so I raise 2 to the power of 4, kept the others the same, and interestingly I get 522*2= 20 total factors. So now I’m unsure as to whether I can eliminate the 20 (Option A).

I suppose my question is, what is the logical way to approach a question like this? And why is the 20 coming up when I raise the 2 to a higher exponent? In your solution video, you raised both 2 and 3 to a power of 2 to get 36 total factors. I, however, decided to change the exponents of 2. Is that wrong?

Thanks,
Q

2

Look at the options first, they consider both positive and negative factors.
To get the number of positive factors alone divide each by two. So we get,
10
12
18
24
27

Now since we have 3 distinct prime divisors, each of them could have any exponent, let’s consider x,y, and z as the three exponents (none of them can be 0), and then the

number of positive factors = (x+1) (y+1) (z+1)

Go back to the options, and see if they can be represented as a product of 3 numbers where each number is greater than or equal to 2.

12 = 2 * 2 * 3 ( x = 1, y = 1, z =2 )
18 = 3 * 3 * 2 (x = 2, y = 2, z = 1 )
24 = 2 * 2 * 6 (x =1, y = 1, z = 5 )
27 = 3 * 3 * 3 (x = 2, y = 2, z = 2 )

10 is the only number that cannot be represented as such.

so 20 is the right answer here.

Hope this helps

3

Hi Joan

Edit: just realized that my example below doesn’t hold because 20 doesn’t factor in the negatives, so 20*2= 40!

And in my original question I actually had a math mistake. In point #6, 20*2= 40! :woman_facepalming:t4::woman_facepalming:t4::woman_facepalming:t4:

I understand it now, thank you for explaining!

How do we know that the following isn’t possible:

20 = 2 * 2 * 5 ( x = 1, y = 1, z =4 )

The exponents can really be any number, can they not?

4

20 includes both positive and negative factors. 10 is the number of positive factors alone.

Then I guess your question is why 10 cannot be represented as (x+1) * (y+1) *(z+1).

It can be only if one of the exponents becomes 0.

i.e

10 = 2 * 5 * 1 (x = 1, y = 4, z = 0)

Any exponent cannot be zero in this case. Therefore, 10 positive factors are not possible for any integer. By extension 20 total factors are also not possible.