Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 21 Oct 2013
Posts: 435

Which of the following cannot be the least common multiple [#permalink]
Show Tags
Updated on: 07 Feb 2014, 05:16
Question Stats:
49% (00:45) correct 51% (00:41) wrong based on 424 sessions
HideShow timer Statistics
Which of the following cannot be the least common multiple of distinct positive integers x and y ? I. 1 II. xy III. x  y A. I only B. III only C. I and II D. I and III E. II and III OE (I): To have LCM = 1, x & y needs to be 1 → But, as x & y are distinct integer → Always No Insufficient
(II): xy is certainly a multiple of x and y, but can it be LCM? Yes, it can be but Not Always If x = 2, y = 3, then xy = 6. → xy = LCM of x & y → 6 is smallest number that is a multiple of 2 & 3. If x = 6, y = 8, then xy = 48 → 24 is LCM As it can be, Sufficient
(III): x and y must be different integers If x > y → (x – y) is positive integer that is less than x. Any multiple of positive integer must be at least equal to or greater than that number. In this case x  y cannot be LCM. If y > x → (x – y) is negative number with absolute value that is less than y. Even though difference is negative, fact that absolute value is less than factor y means that (x – y) cannot be LCM. Insufficient Hi, I want to know if there is other way to solve Statement (III), please.
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by goodyear2013 on 06 Feb 2014, 03:24.
Last edited by Bunuel on 07 Feb 2014, 05:16, edited 2 times in total.
Edited the question.



Intern
Joined: 01 Aug 2006
Posts: 34

Re: Which of the following cannot be the least common multiple [#permalink]
Show Tags
06 Feb 2014, 10:36
1. Because x and y are distinct +ve integers, LCM cannot be 1. (HCF will be 1 if they are coprime). 2. xy may/may not be LCM, but COULD be the LCM. Example: x = 2 and y = 3. LCM = 6 = 2*3. 3. (x  y) can NEVER be the LCM of x and y. If x > y > 0, the lowest possible value of LCM is x. Example: x = 4; y = 2. LCM = 4 = x. If y > x > 0: Example: y = 4; x = 2, x  y = 2; cannot be LCM.
I and III can never be the LCM.



Math Expert
Joined: 02 Sep 2009
Posts: 46286

Re: Which of the following cannot be the least common multiple [#permalink]
Show Tags
07 Feb 2014, 05:17
goodyear2013 wrote: Which of the following cannot be the least common multiple of distinct positive integers x and y ? I. 1 II. xy III. x  y A. I only B. III only C. I and II D. I and III E. II and III OE (I): To have LCM = 1, x & y needs to be 1 → But, as x & y are distinct integer → Always No Insufficient
(II): xy is certainly a multiple of x and y, but can it be LCM? Yes, it can be but Not Always If x = 2, y = 3, then xy = 6. → xy = LCM of x & y → 6 is smallest number that is a multiple of 2 & 3. If x = 6, y = 8, then xy = 48 → 24 is LCM As it can be, Sufficient
(III): x and y must be different integers If x > y → (x – y) is positive integer that is less than x. Any multiple of positive integer must be at least equal to or greater than that number. In this case x  y cannot be LCM. If y > x → (x – y) is negative number with absolute value that is less than y. Even though difference is negative, fact that absolute value is less than factor y means that (x – y) cannot be LCM. Insufficient Hi, I want to know if there is other way to solve Statement (III), please. Similar question to practice: whichofthefollowingcannotbethegreatestcommondivisor108865.html#p866951
_________________
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



Intern
Joined: 21 Aug 2015
Posts: 1

Re: Which of the following cannot be the least common multiple [#permalink]
Show Tags
02 Sep 2015, 22:24
shreyas wrote: 1. Because x and y are distinct +ve integers, LCM cannot be 1. (HCF will be 1 if they are coprime). 2. xy may/may not be LCM, but COULD be the LCM. Example: x = 2 and y = 3. LCM = 6 = 2*3. 3. (x  y) can NEVER be the LCM of x and y. If x > y > 0, the lowest possible value of LCM is x. Example: x = 4; y = 2. LCM = 4 = x. If y > x > 0: Example: y = 4; x = 2, x  y = 2; cannot be LCM.
I and III can never be the LCM. if X=1 and Y=1 then why lcm 1 is not possible.



Intern
Joined: 09 Feb 2014
Posts: 8

Re: Which of the following cannot be the least common multiple [#permalink]
Show Tags
14 Nov 2015, 06:00
anujsabre16@yahoo.com wrote: shreyas wrote: 1. Because x and y are distinct +ve integers, LCM cannot be 1. (HCF will be 1 if they are coprime). 2. xy may/may not be LCM, but COULD be the LCM. Example: x = 2 and y = 3. LCM = 6 = 2*3. 3. (x  y) can NEVER be the LCM of x and y. If x > y > 0, the lowest possible value of LCM is x. Example: x = 4; y = 2. LCM = 4 = x. If y > x > 0: Example: y = 4; x = 2, x  y = 2; cannot be LCM.
I and III can never be the LCM. if X=1 and Y=1 then why lcm 1 is not possible. In the question stem: x and y are distinct positive integers. Thus, x=1 or y=1; not x=1 and y=1 So if x=1, y=2 => LCM is 2 x=2, y=3 => LCM is 6 and so on LCM is 1 only when x=1, y=1



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 11815
Location: United States (CA)
GRE 1: 340 Q170 V170

Re: Which of the following cannot be the least common multiple [#permalink]
Show Tags
03 Dec 2015, 23:51
Hi All, This question is really just a test of your overall understanding of the concept of 'least common multiple.' You don't have to do any fancy math to get the correct answer (and the design of the answer choices helps to avoid a certain amount of thinking/work). The Least Common Multiple between two integers is the smallest number that is a positive multiple of BOTH integers. For example: The LCM of 2 and 3 is 6 The LCM of 2 and 4 is 4 The LCM of 3 and 5 is 15 Notice how the LCM is ALWAYS greater than OR equal to both of the integers involved. Knowing THAT rule, you can quickly deduce which Roman Numerals CANNOT be the LCM of 2 distinct positive integers.... 1) Since the integers are DISTINCT (meaning 'different'), the number 1 CANNOT be the LCM (it would have to be at least 2, and that happens only when the integers are 1 and 2  other LCMs would be bigger). 2) Since the LCM is equal to OR greater than each of the integers, subtracting one integer from the other is NOT going to lead to the LCM (it would lead to a smaller number). These facts allow us to quickly eliminate Roman Numerals 1 and 3. Based on the answer choices, there's no more work to be done. 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: 2570

Re: Which of the following cannot be the least common multiple [#permalink]
Show Tags
08 Feb 2017, 11:34
goodyear2013 wrote: Which of the following cannot be the least common multiple of distinct positive integers x and y ?
I. 1 II. xy III. x  y
A. I only B. III only C. I and II D. I and III E. II and III Let’s analyze each Roman numeral: Roman Numeral I: Can the LCM of two distinct positive integers equal 1? The LCM of two numbers is always greater than or equal to either of the numbers. Even if one of the numbers is 1, the other number must be greater than 1, and therefore, the LCM will be greater than 1. Roman Numeral I cannot be the least common multiple of x and y. Roman Numeral II: Can the LCM of two distinct positive integers equal their product? If two numbers are relatively prime, then their LCM will equal their product. For example, if we take x = 4 and y = 9, then the LCM of x and y is 36, which is equal to their product. Roman Numeral II can be the least common multiple of x and y. Roman Numeral III: Can the LCM of two distinct positive integers equal their difference? Since the LCM of two integers is greater than or equal to either of the numbers, the LCM cannot equal their difference. Roman Numeral III cannot be the least common multiple of x and y. Answer: D
_________________
Jeffery Miller
Head of GMAT Instruction
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions



BSchool Forum Moderator
Joined: 12 Aug 2015
Posts: 2642
GRE 1: 323 Q169 V154

Re: Which of the following cannot be the least common multiple [#permalink]
Show Tags
29 Apr 2017, 17:18
Excellent Question. Here is what i did on this one >
Since the two integers are distinct =>For Least value of LCM =>(x,y)= (1,2) and the least LCM=> 2
Hence LCM≥2 Option 1 > Rejected. This can never be the LCM. So the answer choice must include this one.
Option 2 > LCM would be the product if GCD is one. Eg => (x,y)=(3,5)=> LCM=> 3*5 Hence Acceptable LCM So the answer choice must not contain this.
Option 3 => LCM is always greater than or equal all the involved numbers. Hence as xy is less than x => It can never be the LCM of (x,y) Hence Rejected. So the answer choice must include this option III
=> I and III can never be the LCM
SMASH THAT D.
_________________
MBA Financing: INDIAN PUBLIC BANKS vs PRODIGY FINANCE! Getting into HOLLYWOOD with an MBA! The MOST AFFORDABLE MBA programs!STONECOLD's BRUTAL Mock Tests for GMATQuant(700+)AVERAGE GRE Scores At The Top Business Schools!



Intern
Joined: 27 Apr 2015
Posts: 40

Re: Which of the following cannot be the least common multiple [#permalink]
Show Tags
16 Aug 2017, 17:13
goodyear2013 wrote: Which of the following cannot be the least common multiple of distinct positive integers x and y ? I. 1 II. xy III. x  y A. I only B. III only C. I and II D. I and III E. II and III OE (I): To have LCM = 1, x & y needs to be 1 → But, as x & y are distinct integer → Always No Insufficient
(II): xy is certainly a multiple of x and y, but can it be LCM? Yes, it can be but Not Always If x = 2, y = 3, then xy = 6. → xy = LCM of x & y → 6 is smallest number that is a multiple of 2 & 3. If x = 6, y = 8, then xy = 48 → 24 is LCM As it can be, Sufficient
(III): x and y must be different integers If x > y → (x – y) is positive integer that is less than x. Any multiple of positive integer must be at least equal to or greater than that number. In this case x  y cannot be LCM. If y > x → (x – y) is negative number with absolute value that is less than y. Even though difference is negative, fact that absolute value is less than factor y means that (x – y) cannot be LCM. Insufficient Hi, I want to know if there is other way to solve Statement (III), please. Hi I) LCM (x,y)=1 only if x=y=1. since x,y  distinct +ve integer, LCM≠1 II)LCM=xy Let x=p1^a*p2^b & y=p3^c*p4^d where p1,p2,p3,p4 prime factors and a,b,c,d>=0 Case 1 If p1,p2,p3,p4 distinct prime factors LCM(x,y)=p1^a*p2^b*p3^c*p4^d=xy Case 2 if p1=p3 and a>c LCM(x,y)=p1^a*p2^b*p4^d≠xy Therefore LCM=xy MAY be POSSIBLE III)LCM(x,y)=(xy) (xy) WILL ALWAYS be < Larger of x ,y Since LCM is a multiple of x and y therefore LCM WILL ALWAYS be ≥ Larger of x,y LCM(x,y)≠(xy) Therefore option D Thanks Dinesh




Re: Which of the following cannot be the least common multiple
[#permalink]
16 Aug 2017, 17:13






