Oct 20 07:00 AM PDT  09:00 AM PDT Get personalized insights on how to achieve your Target Quant Score. Oct 22 09:00 AM PDT  10:00 AM PDT Watch & learn the Do's and Don’ts for your upcoming interview 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) Oct 23 08:00 AM PDT  09:00 AM PDT Join an exclusive interview with the people behind the test. If you're taking the GMAT, this is a webinar you cannot afford to miss! Oct 26 07:00 AM PDT  09:00 AM PDT Want to score 90 percentile or higher on GMAT CR? Attend this free webinar to learn how to prethink assumptions and solve the most challenging questions in less than 2 minutes. Oct 27 07:00 AM EDT  09:00 AM PDT Exclusive offer! Get 400+ Practice Questions, 25 Video lessons and 6+ Webinars for FREE. Oct 27 08:00 PM EDT  09:00 PM EDT Strategies and techniques for approaching featured GMAT topics. One hour of live, online instruction
Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 17 Jul 2010
Posts: 109

If x is the product of the positive integers from 1 to 8, in
[#permalink]
Show Tags
21 Jan 2013, 15:11
Question Stats:
78% (01:45) correct 22% (02:15) wrong based on 426 sessions
HideShow timer Statistics
If x is the product of the positive integers from 1 to 8, inclusive, and if i, k, m, and p are positive integers such that x = 2^i * 3^k * 5^m * 7^p, then i + k + m + p = A. 4 B. 7 C. 8 D. 11 E. 12 The OG Guide and MGMAT Guide both have different solutions, a bit long. Can someone tell me if I'm doing this incorrectly. If I'm plugging #'s in, I'm getting 2+3+4+5, = 14, but not all can be added, because not all are prime, and some numbers are repeated right, so if I take the sum of all primes in 2+3+4+5, without repeats I'll get 1+3+2+5, then I get 11? is this correct? I know 1 is not prime, and the first 2, and 4 share the same primes, so do I use 1 as a digit for 2, and use 2 as a prime # for 4? to end up with 1+3+2+5?
Official Answer and Stats are available only to registered users. Register/ Login.



Math Expert
Joined: 02 Sep 2009
Posts: 58434

Re: If x is the product of the positive integers from 1 to 8, in
[#permalink]
Show Tags
21 Jan 2013, 15:29
If x is the product of the positive integers from 1 to 8, inclusive, and if i, k, m, and p are positive integers such that x = 2^i * 3^k * 5^m * 7^p, then i + k + m + p =A. 4 B. 7 C. 8 D. 11 E. 12 Given that \(x=8!=2^7*3^2*5*7\). Hence, \(x=2^7*3^2*5^1*7^1=2^i * 3^k * 5^m * 7^p\), since i, k, m, and p are positive integers, then we can equate the exponents, so we have that \(i=7\), \(k=2\), \(m=1\), and \(p=1\). Therefore, \(i + k + m + p=7+2+1+1=11\). Answer: D. Hope it's clear.
_________________



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9704
Location: Pune, India

Re: If x is the product of the positive integers from 1 to 8, in
[#permalink]
Show Tags
21 Jan 2013, 20:56
laythesmack23 wrote: If x is the product of the positive integers from 1 to 8, inclusive, and if i, k, m, and p are positive integers such that x = 2^i * 3^k * 5^m * 7^p, then i + k + m + p = A. 4 B. 7 C. 8 D. 11 E. 12 The OG Guide and MGMAT Guide both have different solutions, a bit long. Can someone tell me if I'm doing this incorrectly. If I'm plugging #'s in, I'm getting 2+3+4+5, = 14, but not all can be added, because not all are prime, and some numbers are repeated right, so if I take the sum of all primes in 2+3+4+5, without repeats I'll get 1+3+2+5, then I get 11? is this correct? I know 1 is not prime, and the first 2, and 4 share the same primes, so do I use 1 as a digit for 2, and use 2 as a prime # for 4? to end up with 1+3+2+5? You cannot plug in numbers. You need to find the values of i, k, m and p. x = 1*2*3*4*5*6*7*8 = 8! \(x = 2^i*3^k*5^m*7^p\) To get the value of i, you need to find the number of 2s in x i.e. 8! (including the 2s you get in 4, 6 and 8). You can quickly count  one from 2, two from 4, one from 6 and three from 8 = total seven 2s are there in 8! To get the value of k, you need to find the number of 3s in 8!. There are two 3s in 8! (one from 3 and another from 6) It is easy to see that there is only one 5 and one 7 in 8!. \(x = 2^7*3^2*5^1*7^1\) So 7 + 2 + 1 + 1 = 11 Check this post for more on powers in factorials: http://www.veritasprep.com/blog/2011/06 ... actorials/
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Director
Joined: 14 Dec 2012
Posts: 703
Location: India
Concentration: General Management, Operations
GPA: 3.6

Re: If x is the product of the positive integers from 1 to 8, in
[#permalink]
Show Tags
11 Aug 2013, 01:51
Stiv wrote: If x is the product of the positive integers from 1 to 8, inclusive, and if i, k, m and p are positive integers such that \(x = 2^i3^k5^m7^p\), then i + k + m + p = A 4 B 7 C 8 D 11 E 12 \(X= 1*2*3*4*5*6*7*8\) OR =\(1*2*3*2^2*5*(2*3)*7*2^3\) =\(1*2^7*3^2*5*7\) THEREFORE \(i + k + m + p = 7+2+1+1 = 11\) HENCE D
_________________
When you want to succeed as bad as you want to breathe ...then you will be successfull....
GIVE VALUE TO OFFICIAL QUESTIONS...
GMAT RCs VOCABULARY LIST: http://gmatclub.com/forum/vocabularylistforgmatreadingcomprehension155228.html learn AWA writing techniques while watching video : http://www.gmatprepnow.com/module/gmatanalyticalwritingassessment : http://www.youtube.com/watch?v=APt9ITygGss



Intern
Status: Preparation
Joined: 03 Apr 2012
Posts: 6
Location: India
GPA: 2.9

Re: If x is the product of the positive integers from 1 to 8, in
[#permalink]
Show Tags
11 Aug 2013, 02:40
If given any factorial .Then the maximum power of a prime number in that factorial can be obtained by using the following formula.
If given n! then the maximum power of a prime number p in that factorial is obtained by
[n/p] + [n/p^2] + [n/p^3] + ......
where [x] is a step function and gives the greatest integer less than or equal to X.
We have to continue the above formula until we get the value of some term zero.
For example take 8!.
To find the power of 2 in 8!. We apply this formula.
[8/2] + [8/4] + [8/8] + [8/16] +....
= 4 + 2 + 1 + 0 + 0 + ..... = 7



Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2815

Re: If x is the product of the positive integers from 1 to 8, in
[#permalink]
Show Tags
22 Nov 2017, 13:32
feellikequitting wrote: If x is the product of the positive integers from 1 to 8, inclusive, and if i, k, m, and p are positive integers such that x = 2^i * 3^k * 5^m * 7^p, then i + k + m + p =
A. 4 B. 7 C. 8 D. 11 E. 12 Let’s break 8! into prime factors: 8 x 7 x 6 x 5 x 4 x 3 x 2 2^3 x 7 x 2 x 3 x 5 x 2^2 x 3 x 2 2^7 x 3^2 x 5^1 x 7^1 We see that i = 7, k = 2, m = 1, and p = 1. Thus, i + k + m + p = 7 + 2 + 1 + 1 = 11. 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.



SVP
Joined: 26 Mar 2013
Posts: 2345

Re: If x is the product of the positive integers from 1 to 8, in
[#permalink]
Show Tags
20 Mar 2018, 16:24
feellikequitting wrote: If x is the product of the positive integers from 1 to 8, inclusive, and if i, k, m, and p are positive integers such that x = 2^i * 3^k * 5^m * 7^p, then i + k + m + p =
A. 4 B. 7 C. 8 D. 11 E. 12
x is the product of the positive integers from 1 to 8 = 8! = (1) (2) (3) (4) (5) (6) (7) (8) = 2^7 * 3^2 * 5^1 * 7^1 i + k + m + p = 7 + 2+ 1+1 = 11 Answer: D



GMAT Club Legend
Joined: 12 Sep 2015
Posts: 4015
Location: Canada

Re: If x is the product of the positive integers from 1 to 8, in
[#permalink]
Show Tags
30 Dec 2018, 10:44
feellikequitting wrote: If x is the product of the positive integers from 1 to 8, inclusive, and if i, k, m, and p are positive integers such that x = 2^i * 3^k * 5^m * 7^p, then i + k + m + p =
A. 4 B. 7 C. 8 D. 11 E. 12
GIVEN: x = (8)(7)(6)(5)(4)(3)(2)(1) Let's find the prime factorization of x by rewriting each value as the product of primes. We get: x = (2)(2)(2)(7)(2)(3)(5)(2)(2)(3)(2) Rearrange to get: x = (2)(2)(2)(2)(2)(2)(2)(3)(3)(5)(7) Rewrite as powers to get: \(x = (2^7)(3^2)(5^1)(7^1)\) We're told that \(x = (2^i)(3^k)(5^m)(7^p)\) This means: i = 7, k = 2, m = 1 and p = 1 So, i + k + m + p = 7 + 2 + 1 + 1 = 11 Answer: D Cheers, Brent
_________________
Test confidently with gmatprepnow.com




Re: If x is the product of the positive integers from 1 to 8, in
[#permalink]
30 Dec 2018, 10:44






