BTGmoderatorDC wrote:\(n = 2^4*3^2*5^2\) and positive integer d is a divisor of n. Is \(d > \sqrt{n}\) ?
(1) d is divisible by 10.
(2) d is divisible by 36.
OA C
Source: Official Guide
Given that \(n = 2^4*3^2*5^2\), we have \(\sqrt{n} = 2^2*3*5=60\)
Let's take each statement one by one.
(1) d is divisible by 10.
Case 1: Say d = 10 = 2*5, then \(d < \sqrt{n}\). The answer is no.
Case 2: Say \(d = 2^5*3^2*5^2\), then \(d > \sqrt{n}\). The answer is yes.
No unique answer. Insufficient.
(2) d is divisible by 36.
=> d is divisible by \(36 = 2^2*3^2\)
Case 1: Say d = \(36 = 2^2*3^2\), then \(d < \sqrt{n}\). The answer is no.
Case 2: Say \(d = 2^5*3^2*5^2\), then \(d > \sqrt{n}\). The answer is yes.
No unique answer. Insufficient.
(1) and (2) together
From (1) and (2), we see that d is divisible by LCM of 10 and 36, which is \(5*36 = 180 = 2^2*3^2*5> \sqrt{n}\). The answer is yes. Sufficient.
The correct answer:
C
Hope this helps!
-Jay
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