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# How many different positive integers are factors of 441

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

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13 Apr 2012, 02:42
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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?
[Reveal] Spoiler: OA

Last edited by Bunuel on 13 Apr 2012, 06:37, edited 2 times in total.
Edited the question

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

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13 Apr 2012, 04:55
Yeah you are absolutely right!

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
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 factors of 441 [#permalink]

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13 Apr 2012, 06:43
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Expert's post
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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.

Hope it helps.
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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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Kudos [?]: 63 [0], given: 16

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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

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

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15 Aug 2016, 04:28
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Re: How many different positive integers are factors of 441   [#permalink] 15 Aug 2016, 04:28
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