LCM

[maihuna]

If s and t are distinct positive integers, which of the following could NOT be the lowest common positive multiple of s and t?


A : st
B : s/t
C : t
D: 2
E: s+t

[Ossa]

Hi,

There are multiple ways to do this, but here is an easy way: if you choose 1 and 2 as the two integers, then you will realize that the first three answers could give you the LCM, and accordingly the 4th one has to be the right answer.

If s and t are distinct positive integers, which of the following could NOT be the lowest common positive multiple of s and t?


A : st
B : s/t
C : t
D: 2
E: s+t

[logitech]

If s and t are distinct positive integers, which of the following could NOT be the lowest common positive multiple of s and t?


A : st
B : s/t
C : t
D: 2
E: s+t

I love this kind of questions.

S X = LCM

T Y = LCM

So,

ST = SX = TY , we will have X=T and Y=S meaning we should have two numbers who have no common factor, lets say two primes 3 and 5 , and LCM is 3x5=15

S/T = SX = TY, this time, X=1/T , we can only choose 1 as T to make X an integer SO T = 1 and X can be anything, lets say 1 and 3 LCM = 3/1 = 3

T=TY = SX, hmm...Y=1 So T is divided by S lets say, 6 and 2 LCM is 6(t)

2=sx=ty, well how about S=2 and t=1 then we will have LCM as 2



s+t = sx , this actually means that S is a multiple of T or T is a multiple of S and this CAN NOT BE the LCM in my opinion.

I would choose E

Any more takers ?

[cramya]

I am sure thats what Ossa meant s+t when he said 4th he probably meant 4th wiht s and t's. I could be worng also.


Take s=1 t=2

s+t does not work

Done...

[ronniecoleman]

IMO E

choose smart numbers: 1 and 2

[nervesofsteel]

has to be E only

[logitech]

has to be E only

Oh yeah? Interesting...

[vittalgmat]

Thanks Logitech, Ossa for providing insightful explanations.
After looking at your explanations, I realized the I could have picked
2 as one of the "smart numbers". The other number could have been either 1 or 3. Choosing 3 would have limited my options to test while 1 covers more ground.

Just a thought.