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If n=4p, where p is a prime number greater than 2, how many [#permalink]
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23 Jan 2008, 12:07
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If n=4p, where p is a prime number greater than 2, how many different positive even divisors does n have, including n?
A. 2 B. 3 C. 4 D. 6 E. 8
Last edited by blog on 23 Jan 2008, 12:41, edited 1 time in total.



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Re: prime number [#permalink]
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23 Jan 2008, 12:37
Dn=4p=2^2*p N=(2+1)(1+1)=6
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Re: prime number [#permalink]
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23 Jan 2008, 12:42
walker wrote: D
n=4p=2^2*p
N=(2+1)(1+1)=6 Quote: oops m sorry its : where p is a prime no greater than 2 here in the question.



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Re: prime number [#permalink]
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23 Jan 2008, 12:44
The answer is A. The even divisors are 2*p and 4*p. There are two more divisors 1 and p but they are odd since p > 2



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Re: prime number [#permalink]
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23 Jan 2008, 12:47
My pick B (3)
Even divisors  2, 4, and n (n will be even after multiplying with 4)



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Re: prime number [#permalink]
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23 Jan 2008, 12:48
maratikus wrote: The answer is A. The even divisors are 2*p and 4*p. There are two more divisors 1 and p but they are odd since p > 2 but here ans is C.



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Re: prime number [#permalink]
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23 Jan 2008, 12:54
maratikus wrote: The answer is A. The even divisors are 2*p and 4*p. There are two more divisors 1 and p but they are odd since p > 2 I lose " even"... I should go to sleep....byebye!
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Re: prime number [#permalink]
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23 Jan 2008, 12:59
What am I thinking, of course the anser is C:
2, 4, 2*p, and 4*p.
Last two are even numbers too.



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Re: prime number [#permalink]
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23 Jan 2008, 13:00
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that's funny, I totally messed up too  concentration is important.



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Re: prime number [#permalink]
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01 Feb 2008, 03:59
blog wrote: If n=4p, where p is a prime number greater than 2, how many different positive even divisors does n have, including n?
A. 2 B. 3 C. 4 D. 6 E. 8 i plugged an easy number. n=4p n =4*3 n =12 12 = 2^2 * 3 (2+1)(1+1) = total 3*2=6 total  odd = even 6  (2)= 4
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Re: prime number [#permalink]
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01 Feb 2008, 04:51
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I agree with BMW here, why not plug in the easiest numbers you can find? What works in one situation will work for the problem.



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Re: prime number [#permalink]
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01 Feb 2008, 07:06
dont overcomplicate things, just pick numbers n=12,20,28,40... youll see that each of these numbers have 4 even divisors.



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Re: prime number [#permalink]
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01 Feb 2008, 16:16
blog wrote: If n=4p, where p is a prime number greater than 2, how many different positive even divisors does n have, including n?
A. 2 B. 3 C. 4 D. 6 E. 8 (C) 4*p = 2*2*p Hence, 4 even divisors: 2, 4, 2*p, 4*p
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Re: prime number [#permalink]
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01 Feb 2008, 16:37
walker wrote: D
n=4p=2^2*p
N=(2+1)(1+1)=6 nice formulae to remember.
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Re: prime number [#permalink]
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02 Feb 2008, 12:55
GMAT TIGER wrote: walker wrote: D
n=4p=2^2*p
N=(2+1)(1+1)=6 nice formulae to remember. What is the rule for this formula.. ?
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Re: prime number [#permalink]
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02 Feb 2008, 13:08
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neelesh wrote: What is the rule for this formula.. ?
Number of divisors of N = (m+1)(n+1)(p+1)... Where, \(N = (a^m)(b^n)(c^p)..\)and a,b,c.. are prime numbers example, \(24 = 2^3*3^1\); number of divisors = (3+1)(1+1)=8 (1,24,2,4,6,8,3,12)



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Re: prime number [#permalink]
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02 Feb 2008, 13:08
neelesh wrote: What is the rule for this formula.. ? for positive integer \(n=p_1^a*p_2^b...p_m^k\), where \(p_i\)  are prime numbers The number of positive factors is: \(N=(a+1)(b+1)....(k+1)\)
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Re: prime number [#permalink]
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02 Feb 2008, 13:17
what a great formula!!!! that itself will improve my score



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Re: prime number [#permalink]
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02 Feb 2008, 13:46
+1 to both walker and srp. thx.
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