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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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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.

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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]
1
Kudos
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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