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How many different positive integers are factors of 441 [#permalink]
13 Apr 2012, 01:42

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63% (01:45) correct
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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?

Re: How many different positive integers are factor of 441 ? [#permalink]
13 Apr 2012, 05:27

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

Re: How many different positive integers are factors of 441 [#permalink]
13 Apr 2012, 05:43

7

This post received KUDOS

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.

Re: How many different positive integers are factor of 441 ? [#permalink]
14 Apr 2012, 00:00

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

Re: How many different positive integers are factor of 441 ? [#permalink]
14 Apr 2012, 01:13

Expert's post

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

Re: How many different positive integers are factors of 441 [#permalink]
17 Mar 2014, 03:00

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Re: How many different positive integers are factors of 441 [#permalink]
29 May 2015, 08:45

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