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

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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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Updated on: 15 Jul 2016, 05:03
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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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Sentence Correction-Collection of Ron Purewal's "elliptical construction/analogies" for SC Challenges

Originally posted by AbdurRakib on 24 Jun 2016, 12:24.
Last edited by AbdurRakib on 15 Jul 2016, 05:03, edited 2 times in total.
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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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24 Jun 2016, 13:43
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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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14 Oct 2016, 08:19
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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14 Oct 2016, 14:15
9
4
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

C
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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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06 Nov 2016, 23:05
3
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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10 Nov 2016, 23:00
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.

C
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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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25 Jun 2017, 12:28
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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10 Aug 2017, 19:35

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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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28 Aug 2017, 07:02
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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29 Oct 2017, 09:02
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
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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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29 Oct 2017, 09:30
6
1
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
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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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28 Oct 2018, 01:39
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.

Thanks
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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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10 Dec 2018, 05:43
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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11 Dec 2018, 10:17
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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28 Jun 2019, 06:13
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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01 Jul 2019, 10:52
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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05 Sep 2019, 05:50
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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05 Sep 2019, 10:51
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
Re: M and N are integers such that 6<M<N.What is the value of N?   [#permalink] 05 Sep 2019, 10:51
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