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21 Mar 2018, 05:21
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
36% (02:25) correct
64%
(02:28)
wrong
based on 107
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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
(1) D = 1
(2) D is a factor of 3k + 3
21 Mar 2018, 08:36
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Bunuel
\(M = 54k^2 + 93k + 40=54k^2+48k+45k+40=(9k+8)(6K+5)\)..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
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
General Discussion
21 Mar 2018, 05:46
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Bunuel
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
