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How many positive factors?

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by Patrick_GMATFix » Sun Mar 09, 2014 6:43 pm
The number of positive factors follows the formula below:

if x = m^a * n^b * o^c... where m, n and o are the distinct prime factors of x, then x must have (a+1)(b+1)(c+1)... positive factors.

The answer is D. I go through the question in detail in the full solution below (taken from the GMATFix App).

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by vinaycfc » Sun Mar 09, 2014 6:57 pm
How many different positive integers are factors of 441?

A) 4
B) 6
C) 7
D) 9
E) 11


441=21^2

21=7*3

21^2=(7^2*3^2)

num of factors of a^n*b^m=(n+1)*(m+1)

num of factors of 7^2*7^3=(2+1)*(2+1)=3*3=9
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by theCodeToGMAT » Sun Mar 09, 2014 6:57 pm
441 = 7 x 7 x 3 x 3 = (7)^2 * (3)^2

So factors = (2+1)(2+1) = 9
[spoiler]
{D}[/spoiler]
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by [email protected] » Mon Mar 10, 2014 12:23 pm
Hi LulaBrazilia,

This type of question can be answered using a variety of approaches. Here's a method that uses Prime Factorization and a "list" of the options:

441 can be factored down in "any order"; if you can spot that it's 21x21, then great, but you can just factor it down one piece at a time...

441 = 3x147

147 = 3x49

49 = 7x7

So...
441 = 3x3x7x7

Now, let's list the factors. Every "combination" of those 4 numbers is required:
1 --->Don't forget this factor!!!! 1 is a factor of EVERYTHING
3
7
9 = 3x3
21 = 3x7
49 = 7x7
63 = 3x3x7
147 = 3x7x7
441 = 3x3x7x7

Total Factors: D

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by Jeff@TargetTestPrep » Mon Mar 12, 2018 10:16 am
LulaBrazilia wrote:How many different positive integers are factors of 441?

A) 4
B) 6
C) 7
D) 9
E) 11
To determine the total number of positive factors, we first break 441 into its prime factors, add 1 to each exponent and multiply the results.

441 = 7^2 x 3^2

So the number of factors is (2 + 1)(2 + 1) = 3 x 3 = 9.

Answer: D

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