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If n = 4p, where p is a prime number greater than 2 how many

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If n = 4p, where p is a prime number greater than 2 how many [#permalink] New post 14 Aug 2005, 12:43
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A
B
C
D
E

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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?

1. 2
2. 3
3. 4
4. 6
5. 8

2) If n is a positive integer and n^2 is divisible by 72, the the largest +ve integer that must divide n is

1. 6
2. 12
3. 24
4. 36
5. 48


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 [#permalink] New post 14 Aug 2005, 12:54
You should start by breaking n into its prime factors. This means

n = 2*2*p

Since the only even prime number is 2, and we know that p is greater than 2, the only even divisors of n will be

2
2*2=4
2*2*p=n because this is also an even number

I don't know if this qualifies as a shortcut, but it's the solution I can think of.

Vasil

This solution is wrong. See below

Last edited by vasild on 14 Aug 2005, 14:30, edited 1 time in total.
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 [#permalink] New post 14 Aug 2005, 13:05
Thanks, Vasild.

Could you explain the approach to solving the second Q as well?

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 [#permalink] New post 14 Aug 2005, 13:25
Similar approach as above. From the stem we have

n*n = 72*x = 2*2*2*3*3*x, where x is some random number

If we take the square root of both sides, we have

n = 2*3*sqrt(2x)

We know that n must be an integer so sqrt(2x) must be a positive integer. This means that x must be an odd exponent of 2. The lowest possible value of sqrt(2x) is therefore 2.

All that said, n must be a factor of 2*3*2=12, so (B) should be the correct answer.

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 [#permalink] New post 14 Aug 2005, 13:40
I think I get it now :-D ...

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Re: PS - divisors [#permalink] New post 14 Aug 2005, 13:46
both are important questions.
Priti wrote:
1)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


for the first one, suppose p = 4(any prime integer greater than 2, lets suppose 3)= 12

the dividers of 12 are: 1, 2, 3, 4, 6, and 12. altogather 6. so D is the answer.

Priti wrote:
2) If n is a positive integer and n^2 is divisible by 72, the largest +ve integer that must divide n is
a. 6
b. 12
c. 24
d. 36
e. 48


for the second one, n^2 is a square of a positive integer and is also divisible by 72, whose factors are 2x2x2x3x3. so n could be any square number that is divisible by 72. for example 144, 144x4, 144x9, 144x16, 144x25, 144x36. remember n^2 cannot be 72 because n is a positive integer and sqrt(n^2) must be an integer.

lets see with some square values:
if n^2 = 144, n = 12 and n is divisible by 1,2,3,4,6, and 12. so 12 is the largest positive integer.
if n^2 = 144x4, n = 24 and n is divisible by 1,2,3,4,6,8, 12 and 24. so 24 is the largest positive integer. but n^2 could be 144 or 144x4. 24 works for the later one but doesnot work for the former where as 12 works for both so 12 or B is the correct answer.

hope it works.........................
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 [#permalink] New post 14 Aug 2005, 14:29
After reading Himalaya's post I realized that I have made a mistake on the first question.

I considered

2
2*2
2*2*p

as possible divisors, but I omitted 2*p

I think (C) is the correct answer after all.

P.S. Himalaya, we are being asked about the different even divisors, not all divisors.
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Re: PS - divisors [#permalink] New post 14 Aug 2005, 19:26
1)If n = 4p, where p is a prime number greater than 2, how many different positive even divisors does n have, including n?

N=2*2*p (where p is an odd prime number, so can't be separated further)
even divisors:
2, 4, 2p, 4p

2) If n is a positive integer and n^2 is divisible by 72, the the largest +ve integer that must divide n is

n^2=72s=2*2*2*3*3*s =6^2*2s
In other words s=2m, where m is the square of another integer.

The largest positive integer that must divide n can be found when m=1. In other words it is 12.
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 [#permalink] New post 15 Aug 2005, 11:01
late in the game but answers are C and B

1)

just pick numbers...say p =3, or 5...you will notice that you have 4 even factors...

2)
n^2=72...break it down to prime factors
72=2^3. 3^2.

n=3.2^2 (you cant have 2^3/2...) so you need at least 2 2s and one3...12...
  [#permalink] 15 Aug 2005, 11:01
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