Which of the following cannot be the least common multiple

Question

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.

shreyas · Accepted Answer

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.

Bunuel · Answer

Similar question to practice: which-of-the-following-cannot-be-the-greatest-common-divisor-108865.html#p866951

anujsabre16@yahoo.com · Answer

if X=1 and Y=1 then why lcm 1 is not possible.

trambn · Answer

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

EMPOWERgmatRichC · Answer

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:  D

GMAT assassins aren't born, they're made,
Rich

JeffTargetTestPrep · Answer

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

stonecold · Answer

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 and III can never be the LCM

SMASH THAT D.

dineshril · Answer

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

LCM(x,y)≠(x-y)

Therefore option D

Thanks
Dinesh

Kinshook · Answer

Asked: Which of the following cannot be the least common multiple of distinct positive integers x and y ?

LCM(x,y) is a multiple of both x & y

I. 1
1 can not be LCM of x & y since they are distant positive integers and both are not equal to 1; TRUE
II. xy
If x & y are co-prime then LCM(x,y) = xy : FALSE
III. x - y
Since (x-y) have to be multiple of both x & y, it should not be less than x or y, but (x-y) is smaller than x; TRUE

IMO D

BrushMyQuant · Answer

Theory

➡ Larger(a,b) <= LCM (a,b) <= a*b

Which of the following cannot be the least common multiple of distinct positive integers x and y

Let's take each option choice and evaluate

I. 1
As x and y are distinct positive integers so one or both of them will be greater than 1.
=> LCM of these two numbers will be greater than the larger of them => LCM > 1
=> FALSE

II. xy
Now, we can take co-prime values of x and y (co-primes are numbers which have only 1 as as the common factor) and this will be true.
Ex, x=2 and y=3 => LCM = 2*3 = x*y
=> TRUE

III. x - y
Now. LCM ≥ larger of the two numbers x and y
Since, x and y are positive so x-y will be lesser than the larger of x and y
=> LCM cannot be x-y
=> FALSE

So, Answer will be D.
Hope it helps!

Watch the following video to Learn the Basics of LCM and GCD

egmat · Answer

This is a great question to test your understanding of LCM properties. The trick here is that the question asks which CANNOT be the LCM, and you need to remember that x and y are distinct positive integers. Let's work through this together.

Understanding What We Need:

The LCM (least common multiple) of two numbers is the smallest positive number that both original numbers divide into evenly. Here's the key property you need to remember: The LCM of any two positive integers must be greater than or equal to each of the original numbers.

In other words: \(LCM(x,y) \geq x\) and \(LCM(x,y) \geq y\)

Let's Test Each Statement:

Statement I: Can LCM = 1?

Think about this: For the LCM to be 1, both x and y would need to be 1 (since 1 is the only positive integer whose LCM with itself is 1). But the question tells us x and y are distinct positive integers, meaning \(x \neq y\). So this is impossible. ✗

Statement II: Can LCM = xy?

Let's try \(x = 3\) and \(y = 5\):
These are relatively prime (no common factors except 1)
\(LCM(3,5) = 15 = 3 \times 5 = xy\) ✓

This works! When two numbers share no common factors, their LCM equals their product.

Statement III: Can LCM = x - y?

Here's where it gets interesting. Let's consider all possible cases:

Case 1: If x > y
Then \(x - y\) is positive, but notice that \(x - y < x\)
This violates our fundamental property that \(LCM(x,y) \geq x\)
For example: If \(x = 7\) and \(y = 3\), then \(x - y = 4\)
But \(LCM(7,3) = 21\), not 4. We'd need the LCM to be 4, but 4 can't even be a multiple of 7!
Case 2: If x < y
Then \(x - y\) is negative
But the LCM must be positive, so this is impossible.
Case 3: If x = y
Wait, the question says x and y are distinct, so this case doesn't apply!

Therefore, \(x - y\) can NEVER be the LCM. ✗

The Answer:

Statements I and III cannot be the LCM, so the answer is D (I and III).

The Key Insight: Notice how the fundamental constraint that \(LCM(x,y) \geq \text{max}(x,y)\) immediately eliminates \(x - y\) when \(x > y\), and the positivity requirement eliminates it when \(x < y\). The "distinct" constraint eliminates 1 and the \(x = y\) case.

Want to Master This Concept?

You can check out the complete step-by-step solution on Neuron by e-GMAT to understand the systematic framework for analyzing LCM problems and learn the common traps students fall into (like forgetting the "distinct" constraint or misreading "CANNOT be"). You can also explore other GMAT official questions with detailed solutions and process skill breakdowns on Neuron here.

Hope this helps! Let me know if you have any questions.

Which of the following cannot be the least common multiple

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Which of the following cannot be the least common multiple

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