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

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

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New post 21 Sep 2019, 18:31
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\(n = 2^4*3^2*5^2\) 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.


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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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New post 21 Sep 2019, 18:43
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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



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

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New post 21 Sep 2019, 23:07
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
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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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New post 22 Sep 2019, 00:18
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]

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New post 18 Dec 2019, 22:00
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] 18 Dec 2019, 22:00

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

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