Which of the following CANNOT be the least common multiple of two posi

The least common multiple of two positive integers cannot be less than either of them. Therefore, since x - y is less than x, it cannot be the LCM of a x and y.

Answer: D.

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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 answer choice CANNOT be the LCM of 2 distinct positive integers....

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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LCM must be greater than or equal to all the involved numbers.
Hence LCM can never be x-y as it is less than x.

Additionally => LCM(x,y) for x>y must lie in the bound => [ x,x*y]


Smash that D
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Since the difference of x - y is less than x, the quantity x - y can’t be a multiple of x. Thus, it can’t be the least common multiple (LCM) of x and y.

(Note: We think the correct answer is intended to be D for the reason stated above, but choice E is also correct since x + y can’t be the LCM of x and y either. We can prove this by contradiction:

Let’s suppose that x + y is the LCM of x and y. We see that x and y can’t be equal, otherwise either x or y (not their sum) will be the LCM of x and y. Now let’s say that x < y. Since we suppose that x + y is the LCM of x and y, y, the larger of the two numbers, can’t be the LCM of x and y. But the LCM of x and y must be the a multiple of y, so it has to be at least 2y (if it can’t be y). Here is the contradiction: 2y > x + y since y > x. So it’s impossible to have x + y as the LCM of x and y.)

Answer: D
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Got the answer , but I am having a hard time finding examples for option E. Can anybody give examples of 2 integers where adding them leads to their LCM. Tried a couple of pairs couldn't get the answer , hence querying. Thanks.
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So E) is true as well?

I knew that D) must be the answer but somehow could not prove E) as right either.
The other ones were easy elimination!
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2-2=0

0 is a multiple of every number, and the LCM of 2 and 2

The least common multiple of two positive integers cannot be less than either of them. Therefore, since x - y is less than x, it cannot be the LCM of a x and y.

Answer: D.

Similar questions to practice:
https://gmatclub.com/forum/which-of-the- ... 08865.html
https://gmatclub.com/forum/which-of-the- ... 67145.html[/quote]

2-2=0

0 is a multiple of every number, and the LCM of 2 and 2[/quote]

Not sure what you are trying to say there. Care to elaborate? Thanks.[/quote]

Yep - isn't 0 a multiple of every number?

TF, if X=Y ie 2 and 2, 2-2=0 which is then LCM of X=2 and Y=2

2-2=0

0 is a multiple of every number, and the LCM of 2 and 2[/quote]

Not sure what you are trying to say there. Care to elaborate? Thanks.[/quote]

Yep - isn't 0 a multiple of every number?

TF, if X=Y ie 2 and 2, 2-2=0 which is then LCM of X=2 and Y=2[/quote]

The least common multiple of two integers a and b is the smallest positive integer that is divisible by both a and b.
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Theory

➡ Larger(a,b) <= LCM (a,b) <= a*b

Which of the following CANNOT be the least common multiple of two positive integers x and y

Let's take each option choice and evaluate

(A) 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

(B) x
This can be true when one number is x and other number is \(\frac{x}{2}\)
Example: One number is 2 (which is \(\frac{x}{2}\)) and other number is 2*2 = 4 (which is x)
=> TRUE

(C) y
This can be true when one number is y and other number is \(\frac{y}{2}\)
Example: One number is 2 (which is \(\frac{y}{2}\)) and other number is 2*2 = 4 (which is y)
=> TRUE

(D) 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

We don't need to check further, but I am solving to complete the solution.

(E) x + y
Couldn't think of any value of x and y for which LCM(x,y) = x + y
We might have to edit this option choice to 2x or 2y or some valid LCM or need to delete this option choice.
=> FALSE

So, Answer will be D or E.
Hope it helps!

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


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