How many different positive integers are factors of 441 ?

MUST KNOW FOR GMAT:

Finding the Number of Factors of an Integer

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.

For more on these issues check Number Theory chapter of Math Book: math-number-theory-88376.html

BACK TO THE ORIGINAL QUESTION.

How many different positive integers are factors of 441
A. 4
B. 6
C. 7
D. 9
E. 11

Make prime factorization of 441: 441=3^2*7^2, hence it has (2+1)(2+1)=9 different positive factors.

Answer: D.

Hope it helps.
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Yeah you are absolutely right!

The answer is 3^2*7*2

Powers of prime factors adding one on to it and multiply


(2+1)*(2+1)=9

Posted from my mobile device

Could you indicate where you found that method? I have not heard of it and I am curious b/c it may be faster than the way I usually approach prime factorization problems....

Usually check if the number is divisible by 2, then 5, then 3, etc... (7, 11, 13,17, ....)
Here 441 is:
- clearly not divisible by 2 or 5 (not even or ending with a 5 or 0)
- divisible by 3 (sum of the digits is 4+4+1=9 divisible by 3)
441 = 3 * 100 + 141 = 3 * 100 + 120 + 21 = 3 *(100+40+7) = 3 * 147 (I would do the middle step in my head so to speak)
same thing with 147
147 = 120 + 27 = 3 * 49
and 49 = 7^2

so 441 = 3^2 * 7^2

Note that I found those links interesting:
https://www.f1gmat.com/data-sufficiency/ ... ility-rule (Not specific to GMAT)
Jeff Sackmann - specific to GMAT strategies:
https://www.gmathacks.com/mental-math/factor-faster.html
https://www.gmathacks.com/gmat-math/numb ... teger.html
https://www.gmathacks.com/main/mental-math.html (general tricks)

is 441 = 7 * 7 * 2 * 2...i think 441 cant be divided by 2

should it not be 7 * 7 * 3 * 3

GreginChicago is saying exactly the same thing!

Check for a solution here: how-many-different-positive-integers-are-factor-of-130628.html#p1073364
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\(441 = 21^2\)

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

\(= 7^2 . 3^2\)

Prime factors = (2+1) * (2+1) = 9

Answer = D

----ASIDE---------------------

If the prime factorization of N = (p^a)(q^b)(r^c) . . . (where p, q, r, etc are different prime numbers), then N has a total of (a+1)(b+1)(c+1)(etc) positive divisors.

Example: 14000 = (2^4)(5^3)(7^1)
So, the number of positive divisors of 14000 = (4+1)(3+1)(1+1) =(5)(4)(2) = 40

-----ONTO THE QUESTION!!----------------------------

441 = (3)(3)(7)(7) = (3^2)(7^2)
So, the number of positive divisors of 441 = (2+1)(2+1)
= (3)(3)
= 9

Answer: D

Cheers,
Brent

the point here is how to count, forget the number, gmat dose not want us to remember this formular.
441=3^2. 7^2
one factor is, 3, 7
two factor is 3.7, 3^2, 7^2
3 factor is , 3^2.7, 3.7^2
4 factor is 3^2 . 7^2.

total is 9

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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APPROACH #2: List and count
When we scan the answer choices (ALWAYS scan the answer choices before settling on an approach), we see that the biggest value is 11.
This means that, if we were to list and count all possible factors, we'd have to list 11 factors at most, which won't take long.

Let's list the factor in PAIRS
We get: 1 & 441
And 3 & 147
And 7 & 63
And 9 & 49
And 21 & 21 [We'll count only one of these 21's]

So, the factors are {1, 3, 7, 9, 21,49, 63 and 441}
TOTAL = 9

Answer: D

Cheers,
Brent

To count a number's divisors, we prime factorize the number, look only at the exponents in the prime factorization, add 1 to each exponent, and multiply what we get. Here, even if you don't recognize 441 as a perfect square, by summing its digits we can tell it's divisible by 9 (since the sum of the digits is divisible by 9). Since 450 = 9*50, then 441 must equal 9*49, so we have

441 = 9*49 = (3^2)(7^2)

and adding one to each exponent and multiplying, we find that 441 has 3*3 = 9 divisors.

Someone asked above why this method, for counting divisors, works. If you think about what a divisor of

3^2 * 7^2

must look like, it must look like this:

3^a * 7^b

where a can be 0, 1 or 2, and b can be 0, 1 or 2. So we have 3 choices for a, and 3 choices for b, and using the fundamental counting principle we use in most counting questions, we have 3*3 = 9 choices for a and b together. Each distinct choice of a and b gives us a different divisor of 3^2 * 7^2, so we have 9 divisors in total.

441= 3^2*7^2
Hence number of factors=(2+1)*(2+1)= 9
IMO D

Posted from my mobile device

This is a question of factorization ,

If a positive integer N can be written in the form of N=(a^p)(b^q)(c^r).... where a,b,c are distinct prime factors of N and p,q,r are positive integers
The no. of distinct positive factors of N=(p+1)(q+1)(r+1)

Here N = 441 = (3^2)(7^2)
Therefore no. of distinct positive factors =(2+1)(2+1) = 9

Hence D