n = 2^4*3^2*5^2 and positive integer d is a divisor of n. Is d > n^?

n = \(2^4\)*\(3^2\)*\(5^2\) and positive integer d is a divisor of n. Is d > \(\sqrt{n}\)

STATEMENT (1)--d is divisible by 10
since d is a divisor of n
a divisor of n divisible by 10 = 10,20.........,100

if d = 10 and \(\sqrt{n}\) = \(2^2\)*3*5 = 60

then -Is d > \(\sqrt{n}\)?---NO

if d = 100

then -Is d > \(\sqrt{n}\)?---YES

INSUFFICIENT

STATEMENT (2)--d is divisible by 36.
if d = 36

then -Is d > \(\sqrt{n}\)?---NO (since, \(\sqrt{n}\) = 60 )

if d = 72

then -Is d > \(\sqrt{n}\)?---YES

INSUFFICIENT

combining both statements together
we know that d is divisible by 36 and 10
taking LCM --minimum d that divides n = 180
all other divisors divisible by 36 and 10 will be greater than 180

--Is d > \(\sqrt{n}\)?---YES

SUFFICIENT

C is the correct answer