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OG 11th Edition PS#241-Need explanation [#permalink]
 02 Apr 2009, 20:16
Question Stats:
0% (00:00) correct
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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]
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. -pradeep
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Re: OG 11th Edition PS#214-Need explanation [#permalink]
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.
-pradeep Thanks pradeep for the explanation! Learn something today.
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Re: OG 11th Edition PS#241-Need explanation [#permalink]
18 Aug 2009, 23:12
thanks, that makes waaay more sense
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Re: OG 11th Edition PS#241-Need explanation [#permalink]
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]
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]
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) and n=t^2therefore n^2=n*n=t*t*n
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Re: OG 11th Edition PS#241-Need explanation [#permalink]
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]
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]
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]
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.
So, answer is B only.
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Re: OG 11th Edition PS#241-Need explanation
[#permalink]
26 Aug 2009, 11:54
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