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If k is an integer and M and D are the least common multiple and the

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If k is an integer and M and D are the least common multiple and the  [#permalink]

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New post 21 Mar 2018, 05:21
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A
B
C
D
E

Difficulty:

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Question Stats:

34% (02:32) correct 66% (01:53) wrong based on 51 sessions

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Re: If k is an integer and M and D are the least common multiple and the  [#permalink]

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New post 21 Mar 2018, 05:46
Bunuel wrote:
If k is an integer and M and D are the least common multiple and the greatest common factor of 9k + 8 and 6k + 5, respectively, is M = 54k^2 + 93k + 40?

(1) D = 1
(2) D is a factor of 3k + 3



the questions translates to are 9k+8 and 6k+5 co prime
if and only if are two numbers co prime the least common multiple would be the value that is multiplied by two co primes

1) d is 1

sufficient
LCM will be M

2) we know lcm < M

sufficient

(D) imo
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If k is an integer and M and D are the least common multiple and the  [#permalink]

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New post 21 Mar 2018, 06:04
Bunuel wrote:
If k is an integer and M and D are the least common multiple and the greatest common factor of 9k + 8 and 6k + 5, respectively, is M = 54k^2 + 93k + 40?

(1) D = 1
(2) D is a factor of 3k + 3



M*D = (9k+8)*(6k+5)
M = (9k+8)*(6k+5)/D

1. Sufficient
2. 3K+3 - can take different values and yield different results

Answer: A
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Re: If k is an integer and M and D are the least common multiple and the  [#permalink]

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New post 21 Mar 2018, 06:08
aggvipul wrote:
Bunuel wrote:
If k is an integer and M and D are the least common multiple and the greatest common factor of 9k + 8 and 6k + 5, respectively, is M = 54k^2 + 93k + 40?

(1) D = 1
(2) D is a factor of 3k + 3



M*D = (9k+8)*(6k+5)
M = (9k+8)*(6k+5)/D

1. Sufficient
2. 3K+3 - can take different values and yield different results

Answer: A

Kudos keeps motivated


you know that 3k+3 can take different value and yield different results and this is sure that m is not equal to 54k^2 + 93k + 40

M can be 54k^2 + 93k + 40
only if 3k+3 = 1 which is not possibile

i guess B is sufficient
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Re: If k is an integer and M and D are the least common multiple and the  [#permalink]

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New post 21 Mar 2018, 07:27
Hatakekakashi wrote:
aggvipul wrote:
Bunuel wrote:
If k is an integer and M and D are the least common multiple and the greatest common factor of 9k + 8 and 6k + 5, respectively, is M = 54k^2 + 93k + 40?

(1) D = 1
(2) D is a factor of 3k + 3



M*D = (9k+8)*(6k+5)
M = (9k+8)*(6k+5)/D

1. Sufficient
2. 3K+3 - can take different values and yield different results

Answer: A

Kudos keeps motivated


you know that 3k+3 can take different value and yield different results and this is sure that m is not equal to 54k^2 + 93k + 40

M can be 54k^2 + 93k + 40
only if 3k+3 = 1 which is not possibile

i guess B is sufficient


For M = 54k^2 + 93k + 40, 3k+3 need not be equal to 1 rather D has to be a factor of 3k+3. Pls note that 1 is a factor(but not the only) for whatsoever value 3k+3 takes
Hope that clarifies
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Re: If k is an integer and M and D are the least common multiple and the  [#permalink]

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New post 21 Mar 2018, 08:36
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2
Bunuel wrote:
If k is an integer and M and D are the least common multiple and the greatest common factor of 9k + 8 and 6k + 5, respectively, is M = 54k^2 + 93k + 40?

(1) D = 1
(2) D is a factor of 3k + 3



\(M = 54k^2 + 93k + 40=54k^2+48k+45k+40=(9k+8)(6K+5)\)..so the question asks : Is the LCM of these two numbers EQUAL to its product..
Now LCM*HCF= product of the two numbers, so HCF=D=1..
so we have to answer if D is 1?


lets see the statements:-

(1) D = 1
straightway gives us the answer..
suff

(2) D is a factor of 3k + 3.
now is the slight tricky part aggvipul
a) \((3k+3)*3 = 9k+9\) ....so 9k+9 and 9k+8 will have no factors in common except 1 as both are CONSECUTIVE integers
b) \((3k+3)*2 = 6k+6\) ....so 6k+6 and 6k+5 will have no factors in common except 1 as both are CONSECUTIVE integers
so the product (9k+8)(6K+5) wil laso have ONLY 1 as common factor with 3k+3..
But D is a common factor, so ONLY possible value is 1
suff

D
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Re: If k is an integer and M and D are the least common multiple and the  [#permalink]

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New post 28 Mar 2018, 02:18
LCMxHCF = N1xN2

MxD = (9k+8)(6k+5)

Basis above, Rephrasing Q stem:
Is D=1

St.1 Sufficient

St.2 D is a factor of 3k+3
As per factor foundation rule,
D is also a factor of (3k+3)xn, where n is an integer.

So, D is a factor of (6k+6) and (9k+9).

ATS, D is also a factor of (6k+5) and (9k+8), the numbers consecutive to above.

As Consecutive Integers are co-primes, D=1.

Sufficient

Ans D

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Re: If k is an integer and M and D are the least common multiple and the   [#permalink] 28 Mar 2018, 02:18
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