Problem Solving

How many factors does 36^2 (square of 36) have?
A) 2
B) 8
C) 24
D) 25
E) 26

I got it wrong but have the answer. Can someone help with answer and good explanation?

How many factors does 36^2 have?
a.2
b.8
c.24
d.25
e.26
OA:D[spoiler][/spoiler]

To determine the number of positive factors of an integer:

1) Prime-factorize the integer
2) Add 1 to each exponent
3) Multiply

36² = 2^4 * 3^4.
Adding 1 to each exponent and multiplying, we get:
(4+1)*(4+1) = 25 factors.

The correct answer is D.

Here's the reasoning:

To determine how many factors can be created from 36² = 2^4 * 3^4, we need to determine the number of choices we have of each prime factor:
For 2, we can use 2^0, 2^1, 2², 2³, or 2^4, giving us 5 choices.
For 3, we can use 3^0, 3^1, 3², 3³, or 3^4, giving us 5 choices.

To combine the number of choices we have of each prime factor, we multiply:
5*5 = 25 factors.

Similar problems:

https://www.beatthegmat.com/divisors-t85731.html
https://www.beatthegmat.com/all-factors- ... 15019.html

A problem about counting only the ODD factors:
https://www.beatthegmat.com/gmat-loves-f ... 72876.html

thanks Anurag, Brent and GMATGuruNY.

GMATGuruNY - thanks for getting to reasoning, as that's what exactly I was looking for. But, somehow, I am still not able to get step #2 Add 1 to each exponent. When I did this, I considered only 4 choices from 2 and 4 from 3 and so 4*4 = 16 and so couldnt find any matching answer and so just went with 8 (guess )

Is it because 1 is ALWAYS a factor or every number? But 1 is not a prime factor --

may be I am missing something basic...

gmatwarroom wrote:thanks Anurag, Brent and GMATGuruNY.

GMATGuruNY - thanks for getting to reasoning, as that's what exactly I was looking for. But, somehow, I am still not able to get step #2 Add 1 to each exponent. When I did this, I considered only 4 choices from 2 and 4 from 3 and so 4*4 = 16 and so couldnt find any matching answer and so just went with 8 (guess )

Is it because 1 is ALWAYS a factor or every number? But 1 is not a prime factor --

may be I am missing something basic...

72 = 2³ * 3².

The prime-factorization here implies the following:
To create a factor of 72, we can choose from TWO BUCKETS.
The first bucket (2³) contains three 2's.
The second bucket (3²) contains two 3's.

From the 2³ bucket we can choose no 2's, one 2, two 2's, or three 2's, for A TOTAL OF 4 OPTIONS -- ONE MORE than 2's exponent.
From the 3² bucket, we can choose no 3's, one 3, or two 3's, for A TOTAL OF 3 OPTIONS -- ONE MORE than 3's exponent.

In each case, the number of options from each bucket is ONE MORE than the value of the exponent.
It is for this reason that we ADD ONE to each exponent in the prime-factorization and multiply:
Total number of factors of 72 = (3+1)(2+1) = 12.

Sorry GMATGuruNY ... still not clear for me

"From the 2³ bucket we can choose no 2's, one 2, two 2's, or three 2's, for A TOTAL OF 4 OPTIONS -- ONE MORE than 2's exponent. "

From above, I infer, following as options
Option-1 : 2 (i.e. min one 2)
Option-2 : 2*2 (i.e. two 2)
Option-3 : 2*2*2 (i.e. max three 2)

what is 4th option?

gmatwarroom wrote:Sorry GMATGuruNY ... still not clear for me

"From the 2³ bucket there are FOUR OPTIONS:
no 2's at all -- a factor of 72 does not have to include any 2's
one 2
two 2's,
three 2's
A TOTAL OF 4 OPTIONS -- ONE MORE than 2's exponent. "

From above, I infer, following as options
Option-1 : 2 (i.e. min one 2)
Option-2 : 2*2 (i.e. two 2)
Option-3 : 2*2*2 (i.e. max three 2)

what is 4th option?

See the portion above that I highlighted in red:
Option 4: NO 2'S AT ALL

So if the prime-factorization of an integer include 2�, there are 6 options:
Option 1: 2� (meaning NO 2's AT ALL)
Option 2: 2¹
Option 3: 2²
Option 4: 2³
Option 5: 2�
Option 6: 2�
Total options = 5+1 = 6.
One more than 2's exponent.

Brent@GMATPrepNow wrote: Note: In most (in not all cases on the GMAT), factors are limited to positive values. I believe that this question would read: How many positive factors does 36^2 have?

In general, we can say that if N = (p^a)(q^b)(r^c)..., where p, q, r etc are prime numbers, then the total number of positive divisors of N is equal to (a+1)(b+1)(c+1)...

Hello Brent,

Why we need to add 1 to the exponents to get positive values ?
And what if it's not specified to calculate "positive factors" ?