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

hcf*LCM= a*b
statement1: only HCF is mentioned, multiple values are possible NS
Statement 2: Only LCM is mentioned , multiple values are possible NS

combining HCF*LCM= product of M*N
and we know M<N hence we can determine the values.

PS: please let me know if my approach is correct.

The answer is C
6*36=216=m*n
Both m & n are greater than 6
M<N can be satisfied under
Pairs (12,18),(9,24) and (8,27)
But only 12,18 can give gcf 6

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If M and N are among 6, 12, 18 and 36 as well as 6 < M < N then M cannot be either 6 or 36 and N cannot be 6. The only test cases from to use are:

Case 1 - M = 12 , N = 18 , GCD = 6 , LCM = 36
Case 2 - M = 12 , N = 36 , GCD = 12 , LCM = 36
Case 3 - M = 18 , N = 36 , GCD = 18 , LCM = 36

The only case that satisfies the limitations of both statement 1 (GCD) and statement 2 (LCM) is case 1 and therefore N is 18 (answer again is C). Hope this helps explain why N cannot be 36

M>6 and N>M

[1] GCD is 6 :
N= 6 * 3 & M = 6 * 2
N= 6 * 5 & M = 6 * 3

here number can be anything - as long as we multiply the number 6 by any of the prime numbers, the statement 2 will be satisfied

[2] LCM = 36 : 2 * 2 * 3 * 3 - N & M can only be formed with the combination of 2's or 3's
given m & n > 6 so possible values are
N= 2 * 3 * 3 & M = 2 * 2 * 3
N= 2 * 2 * 3 * 3 & M = 2 * 2 * 3
N= 2 * 2 * 3 * 3 & M = 2 * 3 * 3

Hence [1] & [2] individually not sufficient but together they yield the number N = 18 & M = 12

It a Very good question !!!! some claps for GMAC

Lets understand what is being asked.

Here we suppose to find the value of N. a concrete , solid value of N

Statement 1 says GCD of M&N is 6 which means they are multiple of 6 or separated from each other by 6
so M&N can be 12 & 18 , 18 & 24 , 24 & 30 .........and so on (Hence Not sufficient)

Staement 2 says LCM of M&N is 36 means MAX value of N can be 36
so M&N can have value as 9 & 12 , 12 & 18 , 18 & 36 , 9 & 36 , 12 & 36 .(Hence Not sufficient)

On Combining we got 12 & 18 as our final answer because it is common in both.
GCD of 12 & 18 is 6 and LCM of 12 & 18 is 36
No other combination satisfy these condition

Option C is correct choice

Hope that Helps!!!!

Bunuel

Can you please provide an elaborated Solution for this problem?

Would really appreciate it if you could explain in detail as to how to think and approach such tough LCM and GCF problems?

Also, can you link some more SIMILAR QUESTIONS if it is possible.

Thanks in advance.

While your question is to Bunuel, I will try to share an detailed explanation that may help you as well as others.

What is the GCD - It is the product of the common terms with the least power (This is in my words and not a textbook perfect definition )

E.g. If there are two integers 12 and 16
Step-1: Break into prime factors. 12 would be \(2^2*3\) and 16 would be \(2^4\)
Step-2: Identify common prime(s) i.e. 2
Step-3: Pick the common prime with the lowest power i.e. \(2^2\) in our example and that is your GCD
GCD - \(2^2\)

Conceptually GCD is the largest number that can divide the 2 integers in question. Try to find an integer greater than 4 that divides both 12 and 16. You won't be able to find one!

What is LCM - It is the product of the common terms with the highest power (Again this is in my words and not a textbook perfect definition )

In the above example, pick out the highest power of all distinct primes i.e. \(2^4\) and \(3^1\)
Hence LCM would be \(2^4*3^1 = 48\)
Conceptually LCM is the smallest multiple of the 2 integers in the question. Try to find an integer less than 48 that is a multiple of both 12 and 16. You won't be able to find one!

On to the question at hand:-

Info. provided in the question:
1. Both M and N are integers
2. Both are greater than 6
3. N is the greater than M (\(N>M\), E.g. Least values of N could be 8 and that of M could be 7)

We are asked to determine the value of N

Statement-1: GCD (Greatest Common Divisor) of M and N is 6 (or \(2*3\))

This states that 6 will be common to both M and N. Plus, as M and N are greater than 6 there will be other primes too. But this statement does not provide insight into those other primes and hence this statement is insufficient. Let me demonstrate that:

\(M = 2*3*5 = 30\)
\(N = 2*3*7 = 42\)

OR

\(M = 2*3*11 = 66\)
\(N = 2*3*13 = 78\)

Here GCD (M, N) is 6 but N can take different values while staying true to the 3 data points provided by the question stem.

Statement-2: LCM (Least Common Divisor) of M and N is 36 (or \(2^2*3^2\))

Basis the definition of LCM, in which we consider the highest powers of all distinct primes, this statement provides information that 2 and 3 are the only primes carried by M and N. But it does not provide insight into the powers of 2 and 3 specific to M and N. E.g. in the examples below \(2^2\) can be part of M in the first example and also part of N in the very last example, and hence this statement is insufficient.

\(M = 2*3*2 = 12\)
\(N = 2*3*3 = 18\)

\(M = 3*3 = 9\)
\(N = 2*3*2 = 12\)

Combining both statements:

When combining we need to ensure that:

1. \(N>M\) - This one is the key!
2. N and M are greater than 6
3. From statement-1 we know that 6 is common to both M and N
4. From statement-2 we know that 2 and 3 are the only primes carried by M and N and their highest powers are 2 (\(2^2\) and \(3^2\))

\(M = 2*3*2 = 12\)
\(N = 2*3*3 = 18\)

You cannot do the below as it would violate the condition that N>M (pt. 1 i.e. Key).
\(M = 2*3*3 = 18\)
\(N = 2*3*2 = 12\)

Ans C (or \(N = 18\))

While it sounds very simple, but as you can observe that the best approach to solving these questions and math questions in general is to list down the various possibilities in an organized fashion (one below another) in your notebook. The biggest mistakes happen when we ignore the data points provided in the question stem E.g. N>M or N,M>6 in this case.

Hope it helps!

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

Cheers,
Brent

Solution:

Statement One Only:

The greatest common divisor of M and N is 6.

There are at least two possible values for N. For example, M = 12 and N = 18 OR M = 18 and N = 24. Notice that in either case, the GCD(M, N) = 6. Statement one alone is not sufficient.

Statement Two Only:

The least common multiple of M and N is 36.

There are at least two possible values for N. For example, M = 9 and N = 12 OR M = 12 and N = 18. Notice that in either case, the LCM(M, N) = 36. Statement two alone is not sufficient.

Statements One and Two Together:

Since the number of values for N is more “restricted” when we know the LCM of N and M, let’s use statement two to determine all the possible values of N (i.e., the number of cases for values of N and M) first and then determine how many of those cases has GCD(M, N) = 6.

Since LCM(M, N) = 36, both M and N are factors of 36. The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36. However, since both M and N are greater than 6, they can only be from the numbers 9, 12, 18, and 36. Furthermore, since N > M, we have:

1) (M, N) = (9, 12)

2) (M, N) = (9, 36)

3) (M, N) = (12, 18)

4) (M, N) = (12, 36)

5) (M, N) = (18, 36)

Now let’s determine the GCD of the 5 cases above:

1) 3

2) 9

3) 6

4) 12

5) 18

We see that only case 3 gives us 6 as the GCD, and in this case, the value of N is 12. So N must be 12.

Answer: C

DS strategy:

firstly, understand the given info, then dissect the stated question

M and N are both positive integers greater than 6...… most importantly M must be greater than 6.....

can we find a single value for N ?

statement 1)

the GCF between M and N is 6

INFERENCE:
both M and N contain 6 in their prime factorization

we can rewrite

M = 6 n1
N = 6 n2

where n1 and n2 are arbitrary integers, which have no common factors between them... why? since if they had a common factor then 6 would not be the GCF

check for validity: M and N are both positive integers greater than 6...… most importantly M must be greater than 6.....


the min value for M is the following = 6 x 2...… which is 12

valid values for N are the following {6 x 3, 6 x 5, 6 x 7...…

clearly insufficient as we can attain several values for N


statement 2)

LCM between M and N is 36


INFERENCE:
both M and N are factors of 36.....

find some factors of 36 quickly

36 1
18 2
9 4
3 12

if M = 9

N could equal {12, 18 or 36}

we can get different values for N, hence insufficient


1) and 2)

from statement 1) N {6x3 , 6x5, 6x7.... up to infinity}

from statement 2) N {12 , 18, 36}

only value in common is 18

C is our answer

OFFICIAL GMAT EXPLANATION

Arithmetic Properties of numbers

1. Given that the greatest common divisor (GCD) of M and N is 6 and 6 < M < N, then it is possible that M = (6)(5) = 30 and N = (6)(7) = 42. However, it is also possible that M = (6)(7) = 42 and N = (6)(11) = 66; NOT sufficient.

2. Given that the least common multiple (LCM) of M and N is 36 and 6 < M < N, then it is possible that M = (4)(3) = 12 and N = (9)(2) = 18. However, it is also possible that M = (4)(3) = 12 and N = (9)(4) = 36; NOT sufficient.

Taking (1) and (2) together, it follows that 6 is a divisor of M and M is a divisor of 36. Therefore, M is among the numbers 6, 12, 18, and 36. For the same reason, N is among the numbers 6, 12, 18, and 36. Since 6 < M < N, it follows that M cannot be 6 or 36 and N cannot be 6. Thus, there are three choices for M and N such that M < N. These three choices are displayed in the table below, which indicates why only one of the choices, namely M = 12 and N = 18, satisfies both (1) and (2).

Both statements together are sufficient.

Why can't N = 36?
36 is a multiple of 36 and also a multiple of 6.

Video solution from Quant Reasoning starts at 11:20
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vijaykhot

The correct pair of n and m is where none of them are 6 (GCD, obviously) or 36 (LCM).

If m = 36, then the GCD will always be m:

9:36
12:36
18:36

So we need a pair where n is less than 36. Only 12:18 satisfies this.

Posted from my mobile device

(1) The greatest common divisor of M and N is 6
This can lead to the following possibilities
6<6*(2)<6*(3,5,7,11...)
They will all have the same denominators
Hence INSuff

(2) The least common multiple of M and N is 3
it can have the possibilities 6<12<18 0r
6<12<36 or 6<18<36
Hence INsuff

When 1 and 2 combined we get
we get 6<12<18

Hence IMO C

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
Since the HCF of (M,N) = 6, we can say that M,N are multiples of 6. Also its given that 6<M<N. But still we can't find a definite value for N.
Statement 1 alone is insufficient.

(2) The least common multiple of M and N is 36

The possible values of M, N which satisfies 6<M<N are (9,12) ,(12,18) ,(12,36), (18,36) (9,36)

Since different possible values of N are possible , Statement 2 alone is insufficient.

If you combine both statements,It's given that HCF of (M,N) = 6 Therefore we could say M = 6 * a and N = 6* b where a,b are co primes and 1< a<b

Since LCM ( M,N ) = 36, 6*a * b = 36
a * b = 6
a * b = 2*3
where a = 2 and b = 3
So M = 6*2 = 12 and N = 6*3 =18

Since we are able to find a definite value of N, Option C is the correct answer.

Thanks,
Clifin J Francis,
GMAT SME

I made a stupid error here in the understanding of LCM while solving. Hope this helps someone else making the same mistake:

Here, on combining both statements, if the LCM is 36 and HCF (highest common factor) is 6, that means the numbers are 6k and 6z where k and z have no factors in common.

36 = 6x6 -- 6 is already a factor so we are left with the other 6 = 2x3 left.
How can we distribute 2 and 3 between the two numbers such that k and z have no factor in common? take k = 2 and z = 3.

so the numbers are - 12 and 18.

Why can't it be any other number??
Because then the LCM will change.
Say.. take k = 2x5 and z = 3x7 such that HCF of 6k and 6z is still 6 but notice that the LCM now is 6x2x5x3x7 (take the common factors once and the non-common factors all individually).