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

Question

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.

DS32402.01

shridhar786 · Accepted Answer

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

Mo2men · Answer

Does this question belong to the new advanced questions by GMAC?

QuantMadeEasy · Answer

Lets rephrase the question, Is d>60?

(1) d can be 10, 20, 30, 60, 90. Not Sufficient.

(2) d can be 36, 90. Not sufficient.

(1)+(2) d is divisible by 10 and 36 both and the smallest number is 180. Sufficient.

C is correct.

AnirudhChalla · Answer

Yes. It is Q.38 in the new advanced questions by GMAC

Basshead · Answer

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

\sqrt{n} = 2^2 * 3 * 5 = 60

Is  d > 60?

(1) d is divisible by 10.

d can be 10, 20, 30, 70. INSUFFICIENT.

(2) d is divisible by 36.

D can be 36, 72, 108, 180. INSUFFICIENT.

(1&2) For d to be divisible by 36 and 10, d must be divisible by 180.

We can conclude  d > 60 . SUFFICIENT.

Answer is C.

wishmasterdj · Answer

n^1/2 = 2^2 * 3* 5
=60

d should be greater than 60

1) n can be 10, 20, etc. so not sufficient
2) n can be 36, 72, etc, so not sufficient

1+ 2 -> d is multiples of both 10 and 36, therefore d is a multiple of their LCM
LCM (10,36) = 180
180> 60
Hence C

ueh55406 · Answer

Hi  Bunuel

can we check these types of question using prime factorisation?

as in √n=  2^2*3*5

(1) d is divisible by 10. : so of the form: 10(n)

factorisation of 10 will render=  2^2*5  << we don't have a factor of 3

(2) d is divisible by 36: 36(n)

factorisation will give us:  2^2*3^2  << No factor of 5

It's only when we combine the two statement (take the highest power of the common factor) we get the all the three required factor.
  2^2*3^2*5  = 180

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