Oct 16 08:00 PM PDT  09:00 PM PDT EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth $100 with the 3 Month Pack ($299) Oct 19 07:00 AM PDT  09:00 AM PDT Does GMAT RC seem like an uphill battle? eGMAT is conducting a free webinar to help you learn reading strategies that can enable you to solve 700+ level RC questions with at least 90% accuracy in less than 10 days. Sat., Oct 19th at 7 am PDT Oct 20 07:00 AM PDT  09:00 AM PDT Get personalized insights on how to achieve your Target Quant Score. Oct 22 08:00 PM PDT  09:00 PM PDT On Demand for $79. For a score of 4951 (from current actual score of 40+) AllInOne Standard & 700+ Level Questions (150 questions)
Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 27 Mar 2012
Posts: 14

How many different positive integers are factors of 441
[#permalink]
Show Tags
Updated on: 13 Apr 2012, 06:37
Question Stats:
70% (01:10) correct 30% (01:28) wrong based on 635 sessions
HideShow timer Statistics
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?
Official Answer and Stats are available only to registered users. Register/ Login.
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




Math Expert
Joined: 02 Sep 2009
Posts: 58372

Re: How many different positive integers are factors of 441
[#permalink]
Show Tags
13 Apr 2012, 06:43
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.
_________________




Intern
Joined: 18 Sep 2011
Posts: 16
Concentration: Entrepreneurship, International Business
GMAT Date: 11092011

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



Manager
Joined: 06 Feb 2012
Posts: 88
WE: Project Management (Other)

Re: How many different positive integers are factor of 441 ?
[#permalink]
Show Tags
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/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)
_________________
Kudos is a great way to say Thank you...



Manager
Status: I will not stop until i realise my goal which is my dream too
Joined: 25 Feb 2010
Posts: 162

Re: How many different positive integers are factor of 441 ?
[#permalink]
Show Tags
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
_________________
Regards, Harsha Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat Satyameva Jayate  Truth alone triumphs



Math Expert
Joined: 02 Sep 2009
Posts: 58372

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



SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1749
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)

Re: How many different positive integers are factors of 441
[#permalink]
Show Tags
27 Mar 2014, 21:02
\(441 = 21^2\) \(= (7 * 3)^2\) \(= 7^2 . 3^2\) Prime factors = (2+1) * (2+1) = 9 Answer = D
_________________
Kindly press "+1 Kudos" to appreciate



GMAT Club Legend
Joined: 12 Sep 2015
Posts: 4003
Location: Canada

Re: How many different positive integers are factors of 441
[#permalink]
Show Tags
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
_________________
Test confidently with gmatprepnow.com



Director
Joined: 29 Jun 2017
Posts: 929

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



Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2816

Re: How many different positive integers are factors of 441
[#permalink]
Show Tags
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
_________________
5star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews
If you find one of my posts helpful, please take a moment to click on the "Kudos" button.



GMAT Club Legend
Joined: 12 Sep 2015
Posts: 4003
Location: Canada

Re: How many different positive integers are factors of 441
[#permalink]
Show Tags
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
_________________
Test confidently with gmatprepnow.com



GMAT Tutor
Joined: 24 Jun 2008
Posts: 1811

Re: How many different positive integers are factors of 441
[#permalink]
Show Tags
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.
_________________
GMAT Tutor in Toronto
If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com



Manager
Joined: 17 May 2018
Posts: 140
Location: India

Re: How many different positive integers are factors of 441
[#permalink]
Show Tags
13 May 2019, 12:49
441= 3^2*7^2 Hence number of factors=(2+1)*(2+1)= 9 IMO D
Posted from my mobile device




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






