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How many different positive integers are factors of 441 [#permalink]
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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?
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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.
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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! 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



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Re: How many different positive integers are factor of 441 ? [#permalink]
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13 Apr 2012, 06:27
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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/datasufficiency/ ... ilityrule (Not specific to GMAT) Jeff Sackmann  specific to GMAT strategies: http://www.gmathacks.com/mentalmath/factorfaster.htmlhttp://www.gmathacks.com/gmatmath/numb ... teger.htmlhttp://www.gmathacks.com/main/mentalmath.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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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 IntegerFirst 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: mathnumbertheory88376.htmlBACK TO THE ORIGINAL QUESTION. How many different positive integers are factors of 441A. 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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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/datasufficiency/ ... ilityrule (Not specific to GMAT) Jeff Sackmann  specific to GMAT strategies: http://www.gmathacks.com/mentalmath/factorfaster.htmlhttp://www.gmathacks.com/gmatmath/numb ... teger.htmlhttp://www.gmathacks.com/main/mentalmath.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 * 7^2Note that I found those links interesting: http://www.f1gmat.com/datasufficiency/ ... ilityrule (Not specific to GMAT) Jeff Sackmann  specific to GMAT strategies: http://www.gmathacks.com/mentalmath/factorfaster.htmlhttp://www.gmathacks.com/gmatmath/numb ... teger.htmlhttp://www.gmathacks.com/main/mentalmath.html (general tricks) is 441 = 7 * 7 * 2 * 2...i think 441 cant be divided by 2
should it not be 7 * 7 * 3 * 3GreginChicago is saying exactly the same thing! Check for a solution here: howmanydifferentpositiveintegersarefactorof130628.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
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
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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