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

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

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

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.

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!

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.

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

avigutman sir how can i eliminate option e with reasoning approach?

I don't believe that process of elimination is a viable approach to this problem, pdfff. If we understand why a LCM has to be at least as big as the larger of the numbers, we can easily pick out the right answer using that reasoning.