M and N are integers such that 6<M<N.What is the value

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M and N are integers such that 6<M<N .What is the value of N?

(1) The greatest common divisor of M and N is 6
(2) The least common multiple of M and N is 36

OG Q 2017 New Question (Book Question: 297)
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by GMATGuruNY » Sat Jul 30, 2016 2:30 am
AbdurRakib wrote:M and N are integers such that 6<M<N .What is the value of N?

(1) The greatest common divisor of M and N is 6
(2) The least common multiple of M and N is 36
Statement 1:
Case 1: M=6*2=12 and N=6*3=18.
Case 2: M=6*2=12 and N=6*5=30.
Since N can be different values, INSUFFICIENT.

Statement 2:
Case 1 also satisfies Statement 2.
In Case 1, N=18.
Case 3: M=12 and N=36.
Since N can be different values, INSUFFICIENT.

Rule: For any two positive integers x and y, xy = (GCF of x and y)(LCM of x and y).

Statements combined:
MN = GCF * LCM = 6*36 = (2*3)(2*2*3*3) = 2*2*2*3*3*3.
Thus, M and N together must comprise the six prime factors above.
Since 6<M<N, we get the following options:
M=2*3*2=12, with the result that N=2*3*3=18.
M=2*3*3=18, with the result that N=2*2*3=12.
The option in red violates the constraint that M<N.
Implication:
Only the case in green will satisfy both statements and the constraint that 6<M<N.
Thus, N=18.
SUFFICIENT.

The correct answer is C.
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by Needgmat » Sat Jul 30, 2016 11:37 am
Statement 2:
Case 1 also satisfies Statement 2.
In Case 1, N=18.
Case 3: M=12 and N=36.
Since N can be different values, INSUFFICIENT.

Hi GMATGuruNY ,

I didn't get statement 2, can you please explain more.

Many thanks in advance.

Kavin

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by [email protected] » Sat Jul 30, 2016 11:53 am
Hi Kavin,

Mitch's approach focuses on TESTing VALUES (which can be useful on many of the Quant questions that you'll see on Test Day).

From the prompt, we're told that M and N are INTEGERS and that 6 < M < N.

In Fact 2, we're told that the LCM of M and N is 36, so we have to TEST VALUES that fit this particular restriction also...

The first TEST from Fact 1 (M=12 and N=18) also fits the information in Fact 2 (that the LCM is 36). By noticing that, you have an immediate example to work with (and N = 18). We need to determine whether N=18 is always the value of N though, so Mitch came up with another example (M=12, N=36). In that case, the answer to the question is N=36. We now have two different answers to the question (What is the value of N?), so Fact 2 is INSUFFICIENT.

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by GMATGuruNY » Sun Jul 31, 2016 2:25 am
Needgmat wrote:Hi GMATGuruNY ,

I didn't get statement 2, can you please explain more.

Many thanks in advance.

Kavin
The LCM (least common multiple) of x and y is the smallest positive integer divisible by both x and y.
Question stem: What is the value of N?

Statement 2: The LCM of M and N is 36
It's possible that M=12 and N=18, since the smallest positive integer divisible by both 12 and 18 is 36.
It's possible that M=12 and N=36, since the smallest positive integer divisible by both 12 and 36 is 36.
Since different values for N are possible, INSUFFICIENT.
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AbdurRakib wrote:
Fri Jul 29, 2016 1:30 pm
M and N are integers such that 6<M<N .What is the value of N?

(1) The greatest common divisor of M and N is 6
(2) The least common multiple of M and N is 36

OG Q 2017 New Question (Book Question: 297)
Given: M and N are integers such that 6 < M < N

Target question: What is the value of N?

Statement 1: The greatest common divisor of M and N is 6
There are infinitely many pairs of values that satisfy the statement 1.
Here are two cases:
Case a: M = 12 and N = 18. In this case, the answer to the target question is N = 18
Case b: M = 18 and N = 24. In this case, the answer to the target question is N = 24
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: The least common multiple of M and N is 36
There are at least two pairs of values that satisfy the statement 2:
Case a: M = 12 and N = 18. In this case, the answer to the target question is N = 18
Case b: M = 18 and N = 36. In this case, the answer to the target question is N = 36
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that the greatest common divisor of M and N is 6
Statement 2 tells us that the least common multiple of M and N is 36

At this point, there are at least two different approaches we can take...

APPROACH #1: List possible pairs of values that satisfy both statements
The good thing here is that statement 2 indirectly tells us N is less than or equal to 36
We know this because statement 2 tells us that 36 is a multiple of N.
So, let's first list all pairs of values that satisfy statement 2 (as well as the given information):
i) M = 12 and N = 36
ii) M = 18 and N = 36
iii) M = 12 and N = 18
That's it!!

Among these three possible pairs of values, only one pair satisfies statement 1: M = 12 and N = 18
Since it must be the case that M = 12 and N = 18, the target question is N = 18


APPROACH #2: Apply a useful rule
--------ASIDE----------------------
There's a nice rule that says:
(greatest common divisor of x and y)(least common multiple of x and y) = xy
Example: x = 10 and y = 15
Greatest common divisor of 10 and 15 = 5
Least common multiple of 10 and 15 = 30
Notice that these values satisfy the above rule, since (5)(30) = (10)(15)
--------BACK TO THE QUESTION! ----------------------
When we apply the above rule, we see that MN = (6)(36) = 216
In other words, MN = (2)(2)(2)(3)(3)(3)
Since it must be true that 6 < M < N ≤ 36, we can see that our options are very limited.
For example it COULD be the case that M = (2)(2)(2) = 8 and N = (3)(3)(3) = 27, but this pair of values does not satisfy statement 1 or statement 2

It COULD also be the case that M = (2)(2)(3) = 12 and N = (2)(3)(3) = 18, AND this pair of values does satisfy statements 1 and 2
Is there any other pairs of values we can use so that 6 < M < N ≤ 36 and MN = 216?
The answer is no.
So, it must be the case that M = 12 and N = 18, the target question is N = 18

Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

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Brent
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