We are given two integers a and b, such that, 2

Question

We are given two integers a and b, such that, 2 < a < b and b is not a multiple of a. Is the remainder of the division of b by a greater than 1?

(1) The least common multiple of a and b is 42

(2) The greatest common factor of a and b is 2

mau5 · Answer

From F.S 1 , for a=6,b=7, we have the remainder as 1, which is not more than 1. Again, for a=3,b=14, we have a remainder as 2, which is more than 1. Insufficient.

From F.S 2, the integers a and b are of the form 2x and 2y, where x and y are co-primes.

Also, 2y = 2x*q + R, where q is a non-negative integer.

or R = 2(y-qx). As x and y are co-primes, thus y is not equal to qx, for any integral value of q. Thus, (y-qx) will never be zero. And as R is always positive, the value of R will always be more than 1. Sufficient.

Note that as a>2, we can have the first value of a only as 4.

B.

BrushMyQuant · Answer

STAT1  is not sufficient as we can have mutiple cases
case 1 a=6, b=7
now remainder when b is divided by a will be 1 which is NOT greater than 1

case 2 a=6, b=14
now reaminder when b is divided by a will be 2 which is greater than 1

So, STAT1 is NOT sufficient

STAT2  : GCD of a and b is 2 means that both a and b are even numbers
And we know that b is not a mutiple of a
so, in any case reamindder when b is divided by a will be more than 1

case1 a=4, b=6
remaidner will be 2 greater than 1

case 2 a=6, b=10
reaminder will be 4 greater than 1

Also, if you notice then the reaminder is a even number greater than or equal to 2
So STAT2 is SUFFICIENT

So,  Answer will be B 
Hope it helps!

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

https://www.youtube.com/watch?v=7jPV6JD0908

jlgdr · Answer

Alright, as Hamilton Lin says, let's do it!

Statement 1 tells us that the LCM of (a,b) is 42. Therefore, we have that b=7,a=6 remainder is 1, but if b=14 and a=3 remainder is 2, therefore two different answers–> Insufficient.

Statement 2, says that the GCF of (a,b) is 2. Therefore since both are even and b is not a multiple of a the remainder will always be 2, therefore Sufficient

B is the correct answer

Cheers
J

EgmatQuantExpert · Answer

Solution:

We are given:

• ‘ b ’ is not a multiple of ‘ a ’.

• Both, ‘ a ’ and ‘ b ’, are greater than  2  and ‘ b ’ is greater than ‘ a ’.

Statement-1 is “The least common multiple of ‘a’ and ‘b’ is 42”.

42= 2*3*7

‘ a ’ and ‘ b ’ can have multiple values based on prime factorization of  42 .

https://e-gmat.com/blogs/wp-content/uploads/2018/03/rr.jpg

Since the value of remainder may or may not greater than 1. Hence, Statement 1 alone is not sufficient to answer the question.

Statement-2 is “The greatest common factor of ‘a’ and ‘b’ is 2.”

Thus, 
 •  a=2x  and  b=2y , where  x  and  y  are co-prime numbers.

Since ‘ b ’ is greater than ‘ a ’, ‘ b ’ on dividing by ‘ a ’ can be written as:

•  2y= 2x*m + r , where ‘ r ’ is the remainder and ‘ r ’ cannot be ‘ 0 ’ as ‘ b ’ is not a multiple of ‘ a ’.

• Since  2y  is an even number,  (2x*m + r)  should be an even number.

o ‘ r ’ must be even.

 Thus, the least value of ‘ r ’ can be  2 .

Hence, Statement 2 alone is sufficient to answer the question.

Answer: B

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

We are given two integers a and b, such that, 2 < a < b and b is not a

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards