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

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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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Re: M and N are integers such that 6<M<N.What is the value of N?  [#permalink]

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Statement 1: greatest common divisor of M and N is 6. So, M and N are multiple of 6. But, an exact value of N cannot be determined. Insufficient!

Statement 2: LCM of M and N is 36. M can be 9 and N can be 12 or M can be 12 and N can be 18. Multiple possible answer. Insufficient!

Combining 1&2, M and N are multiple of 6 and LCM is 36. So the only possible values on M and N can be 12 and 18 respectively.
Sufficient!

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Re: M and N are integers such that 6<M<N.What is the value of N?  [#permalink]

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2
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.
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Re: M and N are integers such that 6<M<N.What is the value of N?  [#permalink]

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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

OG Q 2017 New Question (Book Question: 297)

Both together

LCM*HCF = MN ( each of m, n are multiple of 6 --- assume m = 6k , k is +ve integer )

N =36*6/ 6k thus N = 36/k , since N is a multiple of 6 then k can only be (1,2,3,6)

if k is 1 thus N= 36, M = 6, if k = 2 thus N= 18 and M = 12 , IF K=3 then N= 12 , M = 18 , IF K = 6 then N= 6 and M = 36) the only option that satisfy the constraint ( 6<M<N) is when K= 2 and N=18 , M=12

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Re: M and N are integers such that 6<M<N.What is the value of N?  [#permalink]

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1
3
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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Re: M and N are integers such that 6<M<N.What is the value of N?  [#permalink]

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Statement 1 tells us that between M and N, 2 and 3 are the lowest factors. However we do not know exactly who has 2 and who has 3; there can also be other factors between them. Insufficient.

Stamtement 2 tells us that 2^2 and 3^2 are the highest factors between M and N. However we do not know whether thats the only factors common between them or that there are lower factors of 2 and 3 between them than 2^2 and 2^3.

Combining both statements we understand that 2 and 3 are the lowest factors and 2^2 and 3^2 are the highest factors. So one of them must be 12 and the other must be 18. Since 6<M<N, N must be 18.

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Re: M and N are integers such that 6<M<N.What is the value of N?  [#permalink]

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14101992 wrote:
Statement 1: greatest common divisor of M and N is 6. So, M and N are multiple of 6. But, an exact value of N cannot be determined. Insufficient!

Statement 2: LCM of M and N is 36. M can be 9 and N can be 12 or M can be 12 and N can be 18. Multiple possible answer. Insufficient!

Combining 1&2, M and N are multiple of 6 and LCM is 36. So the only possible values on M and N can be 12 and 18 respectively.
Sufficient!

Hi 14101992,

According the statement 2, LCM can be 36 as well.
The further constrain is given "by the formula (concept)": LCM*HCF = M*N --> 6*36, which tells that LCM cannot be 36 and so only M=12 and N=18 cen be the answer.

C is correct.

Hope it helps.

Matt
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Re: M and N are integers such that 6<M<N.What is the value of N?  [#permalink]

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Don't understand why N cannot be 36? If somebody can please help

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

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Statement 1 and 2: are clearly NOT SUFFICIENT. Can anyone explain the easiest way how together they are sufficient. I just had a lucky guess 'C' which was correct. Thanks.
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M and N are integers such that 6<M<N.What is the value of N?  [#permalink]

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1
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 Manager  S
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Re: M and N are integers such that 6<M<N.What is the value of N?  [#permalink]

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6
1
M>6 and N>M

 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

 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  &  individually not sufficient but together they yield the number N = 18 & M = 12
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M and N are integers such that 6<M<N.What is the value of N?  [#permalink]

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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

OG Q 2017 New Question (Book Question: 297)

Dear Bunuel, what is your take on this question?

Is there any way we could solve this question faster with a formula or something?

Because I kept thinking about what numbers could fit the statements and doing so took some time.

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Re: M and N are integers such that 6<M<N.What is the value of N?  [#permalink]

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2
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!!!!
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Re: M and N are integers such that 6<M<N.What is the value of N?  [#permalink]

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Statement 1: The GCD of M and N is 6. Therefore, M and N must contain 2*3 and may or may not contain any other number
M = 2 * 3 * 7 (any other number or nothing at all)
N = 2 * 3 * 5 (any other number or nothing at all) --- INSUFFICIENT

Statement 2: The LCM of M and N is 36. Therefore, M and N can take the following forms:
M = 2 * 3 * 2 = 12
N = 2 * 3 * 3 = 18

OR

M = 2 * 3 * 2 = 12
N = 2 * 3 * 2 * 3 = 36

--- INSUFFICIENT

Together (1) & (2)
only 1 possibility ,
M = 2 * 3 * 2 = 12
N = 2 * 3 * 3 = 18

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Re: M and N are integers such that 6<M<N.What is the value of N?  [#permalink]

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matt882 wrote:
14101992 wrote:
Statement 1: greatest common divisor of M and N is 6. So, M and N are multiple of 6. But, an exact value of N cannot be determined. Insufficient!

Statement 2: LCM of M and N is 36. M can be 9 and N can be 12 or M can be 12 and N can be 18. Multiple possible answer. Insufficient!

Combining 1&2, M and N are multiple of 6 and LCM is 36. So the only possible values on M and N can be 12 and 18 respectively.
Sufficient!

Hi 14101992,

According the statement 2, LCM can be 36 as well.
The further constrain is given "by the formula (concept)": LCM*HCF = M*N --> 6*36, which tells that LCM cannot be 36 and so only M=12 and N=18 cen be the answer.

C is correct.

Hope it helps.

Matt

Why can't the LCM be 36?
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Re: M and N are integers such that 6<M<N.What is the value of N?  [#permalink]

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I will try to simplify amins309's explanation to some extent:

Statement 1: The GCD of M and N is 6. Therefore, M and N must contain 2*3 and may or may not contain any other number
M = 2 * 3 * 7 (any other number or nothing at all)
N = 2 * 3 * 5 (any other number or nothing at all) --- INSUFFICIENT

Statement 2: The LCM of M and N is 36. Therefore, M and N can take the following forms:
M = 2 * 3 * 2 = 12 N=2*3*3 = 18
M = 2 * 3 * 3 = 18 N=2*2*3*3 = 36
M = 2*3*3 = 18 N= 2*2*3*3 = 36

--- INSUFFICIENT

Together (1) & (2)

In addition to this we use one more property: LCM*HCF = product of two numbers (M*N)
LCM*HCF = 36*6 = 6*6*6 ---- (1)
now in case M=12 and N= 36 --> M*N = 6*6*6*2 is not equal to (1)
Similar to M = 18 and N=36 ---> M*N = 6*6*6*3

While M = 12 and N = 18 ---> M*N = 6*6*6 = LCM*HCF
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Re: M and N are integers such that 6<M<N.What is the value of N?  [#permalink]

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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.

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Re: M and N are integers such that 6<M<N.What is the value of N?  [#permalink]

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S1: if hcf is 6, let m=6a, n=6b, where a & b relatively prime.
=>6<6a<6b
=>1<a<b
Not sufficient

S2: if lcm is 36, 6ab=36;
=> ab=6,
=> a=2, b=3
=> m=2*hcf, n=3*hcf
=> Not sufficient

with S1+S2;
m=12, n=18
Sufficient
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Re: M and N are integers such that 6<M<N.What is the value of N?  [#permalink]

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Inten21 wrote:
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.

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!
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