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Re: n = 2^4*3^2*5^2 and positive integer d is a divisor of n. Is d > n^? [#permalink]
gmatt1476 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.


DS32402.01


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.
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Re: n = 2^4*3^2*5^2 and positive integer d is a divisor of n. Is d > n^? [#permalink]
Mo2men wrote:
gmatt1476 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.


DS32402.01



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


Yes. It is Q.38 in the new advanced questions by GMAC
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Re: n = 2^4*3^2*5^2 and positive integer d is a divisor of n. Is d > n^? [#permalink]
\(n = 2^4*3^2*5^2\) 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.
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Re: n = 2^4*3^2*5^2 and positive integer d is a divisor of n. Is d > n^? [#permalink]
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
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Re: n = 2^4*3^2*5^2 and positive integer d is a divisor of n. Is d > n^? [#permalink]
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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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Re: n = 2^4*3^2*5^2 and positive integer d is a divisor of n. Is d > n^? [#permalink]
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Re: n = 2^4*3^2*5^2 and positive integer d is a divisor of n. Is d > n^? [#permalink]
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