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# OG 11th Edition PS#241-Need explanation

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OG 11th Edition PS#241-Need explanation [#permalink]

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02 Apr 2009, 20:16
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If the integer n has exactly 3 positive divisors, including 1 and n, how many positive divisors does n^2 have?

(A) 4
(B) 5
(C) 6
(D) 8
(E) 9

I don't understand why answer is B. Book explanation not clear.
Thanks!

Last edited by joyseychow on 29 Apr 2009, 02:35, edited 1 time in total.

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Re: OG 11th Edition PS#214-Need explanation [#permalink]

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02 Apr 2009, 21:27
If an integer is to have just 3 factors including one and itself: it has to be the square of a prime.

So n is the square of a prime = 3^2 = 9
n^2 = 9^2 = 81

81 has 5 factors, 1, 81, 3, 9 and 27.

The same applies to all prime numbers. Hope this helps.

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Re: OG 11th Edition PS#214-Need explanation [#permalink]

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03 Apr 2009, 06:55
pbanavara wrote:
If an integer is to have just 3 factors including one and itself: it has to be the square of a prime.

So n is the square of a prime = 3^2 = 9
n^2 = 9^2 = 81

81 has 5 factors, 1, 81, 3, 9 and 27.

The same applies to all prime numbers. Hope this helps.

Thanks pradeep for the explanation! Learn something today.

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Re: OG 11th Edition PS#241-Need explanation [#permalink]

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18 Aug 2009, 23:12
thanks, that makes waaay more sense

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Re: OG 11th Edition PS#241-Need explanation [#permalink]

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18 Aug 2009, 23:34
If the integer n has exactly 3 positive divisors, including 1 and n, how many positive divisors does n^2 have?

From the question, we can get the divisors are:
1, t, n and n=t^2

therefore, for n^2=n*n, it will have 1, t, n, t*n, and n^2 as the divisors.

thus the answer is B) 5.
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Re: OG 11th Edition PS#241-Need explanation [#permalink]

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19 Aug 2009, 07:06
flyingbunny wrote:
If the integer n has exactly 3 positive divisors, including 1 and n, how many positive divisors does n^2 have?

From the question, we can get the divisors are:
1, t, n and n=t^2

therefore, for n^2=n*n, it will have 1, t, n, t*n, and n^2 as the divisors.

thus the answer is B) 5.

hey can you explain how you got this part (in color)

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Re: OG 11th Edition PS#241-Need explanation [#permalink]

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19 Aug 2009, 07:27
ALD wrote:
flyingbunny wrote:
If the integer n has exactly 3 positive divisors, including 1 and n, how many positive divisors does n^2 have?

From the question, we can get the divisors are:
1, t, n and n=t^2

therefore, for n^2=n*n, it will have 1, t, n, t*n, and n^2 as the divisors.

thus the answer is B) 5.

hey can you explain how you got this part (in color)

$$n^2=n*n$$
and $$n=t^2$$

therefore $$n^2=n*n=t*t*n$$
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Re: OG 11th Edition PS#241-Need explanation [#permalink]

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26 Aug 2009, 03:13
I dont understand the concept that if a no. has 3 positive divisors it is a perfect sq..HOW???WHY??
We know..
most nos. have even no. od divisors.
prime nos. have exactly 2 divisors,1 and itself..

Can someone follow the same reasonung and explain the concept to me?
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Re: OG 11th Edition PS#241-Need explanation [#permalink]

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26 Aug 2009, 05:31
tejal777 wrote:
I dont understand the concept that if a no. has 3 positive divisors it is a perfect sq..HOW???WHY??
We know..
most nos. have even no. od divisors.
prime nos. have exactly 2 divisors,1 and itself..

Can someone follow the same reasonung and explain the concept to me?

Let's see..
the number N has EXACTLY 3 devisors: 1, x and itself N. (1<x<N)
If we devide N by x, we get ... another x. (no 1, no N, no another y - as y is fourth devisor). Any N/x=x
4/2=2 ... 9/3=3...
This x is a prime, coz if not then N would have more then 3 devisors

Hope to pour a light on this issue)

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Re: OG 11th Edition PS#241-Need explanation [#permalink]

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26 Aug 2009, 05:55
There is a formula for the number of positive divisors for x:

N = (1+n)*(1+m).... where, x = a^n*b^m....(a, b - prime numbers)

So, in this question: 3=(1+n)(1+m).... it is only possible when n=2 (or m=2...) and other numbers equals 0.

For x^2 = a^2n*b^2m... and N (for x^2) = (1+2n)(1+2m)... = (1+2*2)*1... = 5.
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Re: OG 11th Edition PS#241-Need explanation [#permalink]

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26 Aug 2009, 11:54
let 1, n1, n are the 3 factors of n, then n^2 will have factors:-

1,n1,n1^2,n,n^2.

Consider n^2 as n * n So, if n1 is a factos of n then , it will also be a factor of n^2. Similarly, we have 2 n's in n^2, so n1^2 will also be a factor of n^2.

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Re: OG 11th Edition PS#241-Need explanation   [#permalink] 26 Aug 2009, 11:54
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