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If n=4p, where p is a prime number greater than 2, how many [#permalink ]
23 Jan 2008, 11:07
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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

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blog on 23 Jan 2008, 11:41, edited 1 time in total.

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D n=4p=2^2*p N=(2+1)(1+1)=6

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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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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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My pick B (3) Even divisors - 2, 4, and n (n will be even after multiplying with 4)

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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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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....bye-bye!

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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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that's funny, I totally messed up too - concentration is important.

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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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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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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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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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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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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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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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what a great formula!!!!

that itself will improve my score

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+1 to both walker and srp. thx.

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