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# How many different positive integers are factors of 441

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Joined: 27 Mar 2012
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How many different positive integers are factors of 441  [#permalink]

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Updated on: 13 Apr 2012, 06:37
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Question Stats:

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How many different positive integers are factors of 441

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?

Originally posted by sugu86 on 13 Apr 2012, 02:42.
Last edited by Bunuel on 13 Apr 2012, 06:37, edited 2 times in total.
Edited the question
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Joined: 02 Sep 2009
Posts: 58372
Re: How many different positive integers are factors of 441  [#permalink]

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13 Apr 2012, 06:43
9
7
sugu86 wrote:
How many different positive integers are factors of 441

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?

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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Re: How many different positive integers are factor of 441 ?  [#permalink]

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13 Apr 2012, 04:55
1
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

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Re: How many different positive integers are factor of 441 ?  [#permalink]

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13 Apr 2012, 06:27
1
sugu86 wrote:
How many different positive integers are factor of 441 ? (A) 4 (B) 6 (C) 7 (D) 9 (E) 11. OA is D. 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?

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:
http://www.f1gmat.com/data-sufficiency/ ... ility-rule (Not specific to GMAT)
Jeff Sackmann - specific to GMAT strategies:
http://www.gmathacks.com/mental-math/factor-faster.html
http://www.gmathacks.com/gmat-math/numb ... teger.html
http://www.gmathacks.com/main/mental-math.html (general tricks)
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Re: How many different positive integers are factor of 441 ?  [#permalink]

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14 Apr 2012, 01:00
GreginChicago wrote:
sugu86 wrote:
How many different positive integers are factor of 441 ? (A) 4 (B) 6 (C) 7 (D) 9 (E) 11. OA is D. 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?

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:
http://www.f1gmat.com/data-sufficiency/ ... ility-rule (Not specific to GMAT)
Jeff Sackmann - specific to GMAT strategies:
http://www.gmathacks.com/mental-math/factor-faster.html
http://www.gmathacks.com/gmat-math/numb ... teger.html
http://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
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Re: How many different positive integers are factor of 441 ?  [#permalink]

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14 Apr 2012, 02:13
harshavmrg wrote:
GreginChicago wrote:
sugu86 wrote:
How many different positive integers are factor of 441 ? (A) 4 (B) 6 (C) 7 (D) 9 (E) 11. OA is D. 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?

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:
http://www.f1gmat.com/data-sufficiency/ ... ility-rule (Not specific to GMAT)
Jeff Sackmann - specific to GMAT strategies:
http://www.gmathacks.com/mental-math/factor-faster.html
http://www.gmathacks.com/gmat-math/numb ... teger.html
http://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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Re: How many different positive integers are factors of 441  [#permalink]

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27 Mar 2014, 21:02
$$441 = 21^2$$

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

$$= 7^2 . 3^2$$

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

Answer = D
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Re: How many different positive integers are factors of 441  [#permalink]

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05 Mar 2018, 17:30
Top Contributor
sugu86 wrote:
How many different positive integers are factors of 441

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

----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
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Re: How many different positive integers are factors of 441  [#permalink]

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08 Mar 2018, 07:50
sugu86 wrote:
How many different positive integers are factors of 441

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?

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
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Re: How many different positive integers are factors of 441  [#permalink]

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12 Mar 2018, 09:49
sugu86 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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Re: How many different positive integers are factors of 441  [#permalink]

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16 Apr 2018, 10:51
Top Contributor
sugu86 wrote:
How many different positive integers are factors of 441

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

Alternatively, we might try LISTING and COUNTING the factors of 441.
We'll list them 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
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Re: How many different positive integers are factors of 441  [#permalink]

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13 May 2019, 07:32
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.
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Re: How many different positive integers are factors of 441  [#permalink]

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13 May 2019, 12:49
441= 3^2*7^2
Hence number of factors=(2+1)*(2+1)= 9
IMO D

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Re: How many different positive integers are factors of 441   [#permalink] 13 May 2019, 12:49
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