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Bunuel
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Bunuel
If w and k are distinct positive integers, do they have any common divisors other than 1 ?

(1) k – w = 3w
(2) k and w are even.

(1) k = 4w k is a multiple of w. Sufficient.
(2) k=2x, w=2y. Both of them have at least one 2 as a factor. Sufficient.

Answer D


1) K=4W
If w and k are distinct positive integers, in this case, w must be the common divisor of K and W, but what if the value of w is 1?

if w=1, k=4, there is no other common divisor of W and K other than 1

for other cases, there will always be w as a common divisor for K and w where w>1

Not sufficient


2) 2 will always be a common divisor for K and W, so sufficient


So, answer should be B. Correct me if I am wrong.

+1 Kudos if you like the post :)

Yep, I missed 1 as a valid option for w.
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Bunuel
If w and k are distinct positive integers, do they have any common divisors other than 1 ?

(1) k – w = 3w
(2) k and w are even.

(1) k = 4w k is a multiple of w. Sufficient.
(2) k=2x, w=2y. Both of them have at least one 2 as a factor. Sufficient.

Answer D


1) K=4W
If w and k are distinct positive integers, in this case, w must be the common divisor of K and W, but what if the value of w is 1?

if w=1, k=4, there is no other common divisor of W and K other than 1

for other cases, there will always be w as a common divisor for K and w where w>1

Not sufficient


2) 2 will always be a common divisor for K and W, so sufficient


So, answer should be B. Correct me if I am wrong.

+1 Kudos if you like the post :)

But it should be D, since he has asked for any divisor other than one, nothing specific.
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[/quote]

But it should be D, since he has asked for any divisor other than one, nothing specific.[/quote]

Hello

The question asks whether w and k have any divisor other than 1 or not. So we need to answer this question with a clear cut YES or a clear cut NO. If a statement gives us a sureshot YES or a sureshot NO answer to this question then that statement will be sufficient to answer this question, else not.

In Statement 1, if w=1, then k=4. In this case there is no other common divisor than 1. But if w=2, then k=8, and there is a common divisor of 2 also. So this statement is NOT sufficient to answer the question asked (since there are different answers of YES or NO for different cases).
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