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

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Difficulty:

25% (low)

Question Stats:

59% (01:43) correct
40% (01:11) wrong based on 133 sessions

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

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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