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(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.

Re: Which of the following cannot be the least common multiple [#permalink]

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06 Feb 2014, 10:36

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1. Because x and y are distinct +ve integers, LCM cannot be 1. (HCF will be 1 if they are co-prime). 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): 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.

Re: Which of the following cannot be the least common multiple [#permalink]

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22 Aug 2015, 14:54

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Re: Which of the following cannot be the least common multiple [#permalink]

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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 co-prime). 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.

Re: Which of the following cannot be the least common multiple [#permalink]

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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 co-prime). 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

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.

Re: Which of the following cannot be the least common multiple [#permalink]

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02 Feb 2017, 00:14

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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
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Re: Which of the following cannot be the least common multiple [#permalink]

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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 x-y 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): 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)=(x-y) (x-y) 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

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