How many different positive integers are factors of 441 ?

Question

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

Can this be solved using the method of (p+1) * (q+1) * (r+1) when the given no is of the prime no of the form N=a^p b^q c^r where a,b,c are prime factors ? Can you please clarify whether my approach is correct?

Bunuel · Accepted Answer

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.

sweetzishh5986 · Answer

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

GreginChicago · Answer

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)

harshavmrg · Answer

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

should it not be 7 * 7 * 3 * 3

Bunuel · Answer

GreginChicago is saying exactly the same thing! Check for a solution here: how-many-different-positive-integers-are-factor-of-130628.html#p1073364

PareshGmat · Answer

441 = 21^2

= (7 * 3)^2

= 7^2 . 3^2

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

Answer = D

BrentGMATPrepNow · Answer

----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

thangvietnam · Answer

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

JeffTargetTestPrep · Answer

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

BrentGMATPrepNow · Answer

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  one  of these 21's]

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

Answer: D

Cheers,
Brent

IanStewart · Answer

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.

sumi747 · Answer

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

Posted from my mobile device

Paras96 · Answer

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

PrabhatKC · Answer

So if the question asked for all positive and negative factors of 441, then we multiply the final answer by 2?
Thanks!

Bunuel · Answer

________________________________________
Yes.

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How many different positive integers are factors of 441 ?

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How many different positive integers are factors of 441 ?

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