EgmatQuantExpert wrote:
Bunuel wrote:
Bumping for review and further discussion.
Hi Everyone!
Here's a question to further test your understanding of the concept of GCD:
If A and B are distinct positive integers greater than 1 such that the GCD of A and B is A, then which of the following must be true?
(A) A is a prime number
(B) A and B have the same prime factors.
(C) A and B have the same even-odd nature
(D) All the factors of B are divisible by A
(E) The LCM of A and B is BWill post the solution in this thread on May 1, 2015. Till then, happy solving!
Regards
Japinder
The correct answer is Option E.PFB the correct solution for this question:
We are given that A and B are distinct positive integers greater than 1 such that the GCD of A and B is A
The important thing to note is that the question is asking about
must be true statements. Must be true statements are those that will hold for all possible values of A and B, without exception.
So, our approach here will be to see if we can find any exceptions to the 5 given statements. Let's see.
(A) A is a prime numberConsider A = 20 and B = 60. In this case, GCD(A,B) = A but A is not a prime number. Since we have found an exception to Statement A, it is clearly not a must be true statement.
(B) A and B have the same prime factors.Once again, consider the case of A = 20 and B= 60. The prime factors of A are 2 and 5. The prime factors of B are 2, 3 and 5. So, clearly Statement B doesn't hold true for all possible values of A and B, and therefore, cannot be a must be true statement.
(C) A and B have the same even-odd natureConsider A = 3 and B = 6. Here too, GCD(A,B) = A but the even-odd nature of A and B is opposite. So, Statement C is ruled out as well.
(D) All the factors of B are divisible by A In the case of A= 20 and B = 60, 15 is a factor of B that is not divisible by A.
Similarly, in the case of A = 3 and B = 6, 1 is a factor of B that is not divisible by A
The existence of these exceptions indicates that Statement D is not a must be true statement.
(E) The LCM of A and B is BWe know that LCM(A,B)*GCD(A,B) = A*B . . . (1)
Given: GCD(A,B) = A . . . (2)
On substituting (2) in (1), we get:
LCM(A,B) = B
Therefore, Statement E will always be true, for all values of A and B.
Thanks and Best Regards
Japinder
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