|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
26 May 2017, 04:42
Kudos
Bookmarks
If \(x\) and \(y\) are prime numbers, what is the number of factors of \(18xy\)?
(1) \(x\) and \(y\) are different
(2) \(3 < x < y\)
(1) \(x\) and \(y\) are different
(2) \(3 < x < y\)
26 May 2017, 04:42
Kudos
Bookmarks
Official Solution:
You can take the 1st step of the variable approach and modify the original condition and the question. Whenever you see factors and prime factors, the word "different" is important. In other words, from \(18xy=(2)(3^{2})xy\), \(x\) and \(y\) are different, and also they are also different from 2 and 3. What satisfies this is...
If you take a look at con 2), you can see that from \(3 < x < y\), it satisfies that they are all different.
From \(18xy = (2^{1})(3^{2})(x^{1})(y^{1})\), the number of factors is \((1+1)(2+1)(1+1)(1+1) = 24\), hence it is unique and sufficient. However, in the case of con 1), \(x\) and \(y\) could be either 2 or 3. For example, if \(x = y = 2\), the number of factors from \(18xy=18(2)(2)=(2^{3})(3^{2})\) is \((3+1)(2+1) = 12\), but if \(x = 5\) and \(y = 7\), the number of factors from \(18xy=(2^{1})(3^{2})(5^{1})(7^{1})\) is \((1+1)(2+1)(1+1)(1+1) = 24\), and hence it is not unique and not sufficient. Therefore, the answer is B.
Answer: B
You can take the 1st step of the variable approach and modify the original condition and the question. Whenever you see factors and prime factors, the word "different" is important. In other words, from \(18xy=(2)(3^{2})xy\), \(x\) and \(y\) are different, and also they are also different from 2 and 3. What satisfies this is...
If you take a look at con 2), you can see that from \(3 < x < y\), it satisfies that they are all different.
From \(18xy = (2^{1})(3^{2})(x^{1})(y^{1})\), the number of factors is \((1+1)(2+1)(1+1)(1+1) = 24\), hence it is unique and sufficient. However, in the case of con 1), \(x\) and \(y\) could be either 2 or 3. For example, if \(x = y = 2\), the number of factors from \(18xy=18(2)(2)=(2^{3})(3^{2})\) is \((3+1)(2+1) = 12\), but if \(x = 5\) and \(y = 7\), the number of factors from \(18xy=(2^{1})(3^{2})(5^{1})(7^{1})\) is \((1+1)(2+1)(1+1)(1+1) = 24\), and hence it is not unique and not sufficient. Therefore, the answer is B.
Answer: B
Moderator:
