January 22, 2019 January 22, 2019 10:00 PM PST 11:00 PM PST In case you didn’t notice, we recently held the 1st ever GMAT game show and it was awesome! See who won a full GMAT course, and register to the next one. January 26, 2019 January 26, 2019 07:00 AM PST 09:00 AM PST Attend this webinar to learn how to leverage Meaning and Logic to solve the most challenging Sentence Correction Questions.
Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 21 Oct 2013
Posts: 419

Given that N = a^3b^4c^5 where a, b and c are distinct prime
[#permalink]
Show Tags
Updated on: 28 Feb 2014, 20:43
Question Stats:
69% (01:28) correct 31% (01:40) wrong based on 483 sessions
HideShow timer Statistics
Given that \(N = a^3b^4c^5\) where a, b and c are distinct prime numbers, what is the smallest number with which N should be multiplied such that it becomes a perfect square, a perfect cube as well as a perfect fifth power? A. \(a^3b^4c^5\) B. \(a^5b^4c^3\) C. \(a^2b^3c^5\) D. \(a^7b^6c^5\) E. \(a^{27}b^{26}c^{25}\) While this problem on the surface is an exponent problem, it turns out to be even more of a divisibility, leastcommonmultiple problem. To make N a perfect square, powers of a, b and c must be even; to make N a perfect cube, the powers must be multiples of 3; to make N a perfect fifth power, they must be multiples of 5. This implies that each of the powers of a, b and c must be a multiple of 30 (because 30 is the LCM of 2, 3 and 5) and since we are looking for the smallest number, the powers should each be 30. Hence we need to multiply N by a^27b^26c^25 to make the powers of each of a, b and c equal to 30. Hi, I want to know how I can recognise that question is asking LCM please.
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by goodyear2013 on 27 Feb 2014, 11:09.
Last edited by MacFauz on 28 Feb 2014, 20:43, edited 2 times in total.
Renamed the topic and edited the question.




Math Expert
Joined: 02 Sep 2009
Posts: 52390

Given that N = a^3b^4c^5 where a, b and c are distinct prime
[#permalink]
Show Tags
28 Feb 2014, 00:20
goodyear2013 wrote: Given that N = a^3b^4c^5 where a, b and c are distinct prime numbers, what is the smallest number with which N should be multiplied such that it becomes a perfect square, a perfect cube as well as a perfect fifth power? A. a^3b^4c^5 B. a^5b^4c^3 C. a^2b^3c^5 D. a^7b^6c^5 E. a^27b^26c^25 While this problem on the surface is an exponent problem, it turns out to be even more of a divisibility, leastcommonmultiple problem. To make N a perfect square, powers of a, b and c must be even; to make N a perfect cube, the powers must be multiples of 3; to make N a perfect fifth power, they must be multiples of 5. This implies that each of the powers of a, b and c must be a multiple of 30 (because 30 is the LCM of 2, 3 and 5) and since we are looking for the smallest number, the powers should each be 30. Hence we need to multiply N by a^27b^26c^25 to make the powers of each of a, b and c equal to 30. Hi, I want to know how I can recognise that question is asking LCM please. A perfect square, is an integer that can be written as the square of some integer. For example 16=4^2, is a perfect square. Notice that the powers of the primes of a perfect square are even. A perfect cube, is an integer that can be written as the cube of some integer. For example 27=3^3, is a perfect cube. Notice that the powers of the primes of a perfect cube are multiples of 3. A perfect fifth power, is an integer that can be written as the fifth power of some integer. For example 32=2^5, is a perfect fifth. Notice that the powers of the primes of a perfect fifth power are multiples of 5. According to the above, a number to be a perfect square, a perfect cube as well as a perfect fifth power, must have its primes in power of 30 (the least common multiple of 2, 3, and 5), so it must be a perfect 30th power. The least positive integer \(a^3b^4c^5\) should be multiplied by, in order the powers to be multiples of 30 is \(a^{27}b^{26}c^{25}\): \((a^3b^4c^5)(a^{27}b^{26}c^{25})=a^{30}b^{30}c^{30}=(abc)^{30}\). Answer: E.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




Retired Moderator
Joined: 20 Dec 2013
Posts: 172
Location: United States (NY)
GMAT 1: 640 Q44 V34 GMAT 2: 710 Q48 V40 GMAT 3: 720 Q49 V40
GPA: 3.16
WE: Consulting (Venture Capital)

Re: Given that N = a^3b^4c^5 where a, b and c are distinct prime
[#permalink]
Show Tags
28 Feb 2014, 20:31
oh wow...I spent a good amount of time on this one yesterday before finally getting a headache and hoping Bunuel or Karishma would chime in...I now realize I was attempting to solve a completely different problem As it is formatted, it looks like the question (and relative format of the answer choices) is: N = a^(3b)^(4c)^5 and not: N = (a^3)(b^4)(c^5) many thanks Bunuel!
_________________
MY GMAT BLOG  ADVICE  OPINIONS  ANALYSIS



SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1823
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)

Re: Given that N = a^3b^4c^5 where a, b and c are distinct prime
[#permalink]
Show Tags
03 Mar 2014, 03:12
We require to find LCM of 2, 3 & 5 which is 30... as it should be perfect square AND cube AND Fifth So, 30 should be the power of each expression a, b, & c Answer = E (303) (304 (305)
_________________
Kindly press "+1 Kudos" to appreciate



Manager
Joined: 11 Nov 2011
Posts: 63
Location: United States
Concentration: Finance, Human Resources
GPA: 3.33
WE: Consulting (NonProfit and Government)

Re: Given that N = a^3b^4c^5 where a, b and c are distinct prime
[#permalink]
Show Tags
02 Aug 2015, 06:03
I was intimidated by the exponential when I started reading this question. As it turns out, it is an LCM question.



CEO
Joined: 20 Mar 2014
Posts: 2636
Concentration: Finance, Strategy
GPA: 3.7
WE: Engineering (Aerospace and Defense)

Given that N = a^3b^4c^5 where a, b and c are distinct prime
[#permalink]
Show Tags
02 Aug 2015, 06:18
evdo wrote: I was intimidated by the exponential when I started reading this question. As it turns out, it is an LCM question. Even if you don not recognize that it is an LCM question, you can solve it in 3045 seconds as shown below: \(N = a^3b^4c^5\) Now you need to find more powers of a,b,c such that the 5th root, 3rd root and the square root are all integers (this is what the question means by "such that it becomes a perfect square, a perfect cube as well as a perfect fifth power?" ) For this to happen, N needs to be multiplied by another number (= the correct option to the question asked). Given that \(N = a^3b^4c^5\) > Start with the 5th power or root (as you will see, this is the most restrictive of the 3 required roots). For a number to have an integer value of a "n"th root, it needs to be of the form : \(N = a^n\) > \(N^{1/n} = a\) (if you have any other power then you may not get an integer value when you take the 'n' th root.) Applying the above logic to the question at hand, \(N = a^3b^4c^5\) , we need to make powers of a,b,c multiple of 5 when these options are multiplied by N. Also \(x^a * x^b = x^{a+b}\) Thus, Options A and B are out as they will not provided power of 'a' as a multiple of 5 when these options are multiplied by N. No need to check the other powers for these 2 options. We have options CE remaining. check now for 3rd root or cube power!! For this we need to make sure that the powers of a,b,c are multiples of 3 when these options are multiplied by N. Options C and D are thus eliminated as they do not provide power of 'a' as a multiple of 3 when these options are multiplied by N. E is thus the final answer. Although, this text looks intimidating, it took me not more than 30 seconds to figure out the correct answer.



SVP
Joined: 06 Nov 2014
Posts: 1877

Re: Given that N = a^3b^4c^5 where a, b and c are distinct prime
[#permalink]
Show Tags
15 Jun 2016, 20:27
We need to turn N = a³b⁴c⁵ into a perfect square, cube and a fifth power. Hence whatever are the powers of a, b and c, they should all the divisible by 2, 3, and 5 Or in other words, the powers should be the LCM of 2, 3 and 5 = 30 (Since we need the smallest number)
Hence the multiplication factor should be: a²⁷b²⁶c²⁵
Correct Option: E



Intern
Joined: 22 Jun 2016
Posts: 48

Re: Given that N = a^3b^4c^5 where a, b and c are distinct prime
[#permalink]
Show Tags
09 Jul 2016, 10:22
i dont get this question at all:(( what like which concept is being used here? the question is confusing like what am i supposed to find and how? how is this LCM? i dont get what this is
here is my understanding of the question:
N= a raise to the power 3, b raise to the power 4 and c raise to the power 5
A, B and C are all PRIME numbers. Then out of nowhere, whats the SMALLEST NUMBER with which N should be multiplied? I just dont get it.
so im doing something like N multiply by x (smallest number ) = a number that is perfect sq, perfect cube, perfect fifth power? like where the hell do i go from here? and none of the options give that NUMBER that we need to find. How do i understand this its soo confusing?!!



Retired Moderator
Joined: 19 Mar 2014
Posts: 937
Location: India
Concentration: Finance, Entrepreneurship
GPA: 3.5

Re: Given that N = a^3b^4c^5 where a, b and c are distinct prime
[#permalink]
Show Tags
08 Jul 2017, 13:15
goodyear2013 wrote: Given that \(N = a^3b^4c^5\) where a, b and c are distinct prime numbers, what is the smallest number with which N should be multiplied such that it becomes a perfect square, a perfect cube as well as a perfect fifth power? A. \(a^3b^4c^5\) B. \(a^5b^4c^3\) C. \(a^2b^3c^5\) D. \(a^7b^6c^5\) E. \(a^{27}b^{26}c^{25}\) While this problem on the surface is an exponent problem, it turns out to be even more of a divisibility, leastcommonmultiple problem. To make N a perfect square, powers of a, b and c must be even; to make N a perfect cube, the powers must be multiples of 3; to make N a perfect fifth power, they must be multiples of 5. This implies that each of the powers of a, b and c must be a multiple of 30 (because 30 is the LCM of 2, 3 and 5) and since we are looking for the smallest number, the powers should each be 30. Hence we need to multiply N by a^27b^26c^25 to make the powers of each of a, b and c equal to 30. Hi, I want to know how I can recognise that question is asking LCM please. Lets understand some fundamentals: 1. In order to make an integer a perfect square, we would need the power to be  Multiples of Two
2. In order to make an integer a perfect cube, we would need the power to be  Multiple of Three
5. In order to make an integer a perfect fifth power, we would need the power to be  Multiple of FiveConsidering all of the above, the power should have the LCM of \(2 * 3 * 5 = 30\) So we are looking for  \(N = a^{30}b^{30}c^{30}\) We already have  \(N = a^3b^4c^5\) We would N to be multiplied by  \(N = a^{303}b^{304}c^{305}\) = \(a^{27}b^{26}c^{25}\) Hence, Answer is E
_________________
"Nothing in this world can take the place of persistence. Talent will not: nothing is more common than unsuccessful men with talent. Genius will not; unrewarded genius is almost a proverb. Education will not: the world is full of educated derelicts. Persistence and determination alone are omnipotent."
Best AWA Template: https://gmatclub.com/forum/howtoget60awamyguide64327.html#p470475



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 13375
Location: United States (CA)

Re: Given that N = a^3b^4c^5 where a, b and c are distinct prime
[#permalink]
Show Tags
23 Jan 2018, 12:39
Hi All, This question is built around a few Number Properties involving exponents. For a number to be a perfect square, all of the "exponent terms" must be EVEN. for example…. 25 is a perfect square because 25 = 5^2 16 is a perfect square because 16 = 4^2 = 2^4 For a number to be a perfect cube, all of the "exponent terms" must be A MULTIPLE OF 3. 8 is a perfect cube because 8 = 2^3 64 is a perfect cube because 64 = 4^3 = 2^6 For a number to be a perfect fifth power, all of the "exponent terms" must be A MULTIPLE OF 5. 32 is a perfect fifth power because 32 = 2^5 1024 is a perfect fifth power because 1024 = 4^5 = 2^10 Here, we need each exponent to become a multiple of 2, 3 and 5. The prompt refers to the SMALLEST number, so we need the Least Common Multiple of 2, 3 and 5 ……which is 30. The correct answer will multiply by the given prompt to equal 30. Since we're starting with (A^3)(B^4)(C^5), we'll need to multiply with something that will end with (A^30)(B^30)(C^30). Final Answer: GMAT assassins aren't born, they're made, Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com
Rich Cohen
CoFounder & GMAT Assassin
Special Offer: Save $75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/
*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****



Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2830

Re: Given that N = a^3b^4c^5 where a, b and c are distinct prime
[#permalink]
Show Tags
03 May 2018, 15:25
goodyear2013 wrote: Given that \(N = a^3b^4c^5\) where a, b and c are distinct prime numbers, what is the smallest number with which N should be multiplied such that it becomes a perfect square, a perfect cube as well as a perfect fifth power?
A. \(a^3b^4c^5\) B. \(a^5b^4c^3\) C. \(a^2b^3c^5\) D. \(a^7b^6c^5\) E. \(a^{27}b^{26}c^{25}\) The exponents of a perfect square are all multiples of 2. The exponents of a perfect cube must be all multiples of 3. And the exponents of a perfect fifth power must be all multiples of 5. Since we need N to be a perfect square, perfect cube, and perfect fifth power, we need each exponent to be a multiple of the LCM of 2, 3, and 5, which is 30. Hence, we know that the smallest number of each exponent must be 30. The original exponent of a is 3; so we need 27 more a’s to get to 30: a^3 x a^27 = a^30 The original exponent of b is 4; so we need 26 more b’s to get to 30: b^4 x b^26 = b^30 The original exponent of c is 5; so we need 25 more c’s to get to 30: c^5 x c^25 = c^30 Therefore, we need to multiply N = a^3b^4c^5 by a^27 x b^26 x c^25 to make it become a perfect square, a perfect cube and a perfect fifth power. Answer: E
_________________
Jeffery Miller
Head of GMAT Instruction
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions




Re: Given that N = a^3b^4c^5 where a, b and c are distinct prime &nbs
[#permalink]
03 May 2018, 15:25






