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If n is a positive integer and n^2 is divisible by 72, then

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If n is a positive integer and n^2 is divisible by 72, then [#permalink] New post 08 Jan 2007, 23:21
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A
B
C
D
E

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2. If n is a positive integer and n^2 is divisible by 72, then the largest possible positive integer that must divide n is
A) 6
B) 12
C) 24
D) 36
E) 48
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 [#permalink] New post 09 Jan 2007, 07:53
I get E on this one. Its a pain because I started with 12, the easy answer. 12 square is 144/72=2. 36 works but 48 square is 2304/72 is 32. Hence E. I dont know my squares that high, took a minute to do it longhand. Hope not to see it on the test.

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 [#permalink] New post 09 Jan 2007, 08:06
(B) for me :)

72 = 9*8 = (3)^2 * (2)^2 * 2

n^2 = ( (3)^2 * (2)^2 * 2 ) * k^2
=> n = 3*2*sqrt(2)*k : non sense, as n is an integer, it forces k to not be an integer.

We must have an integer j such that :
n^2 = ( (3)^2 * (2)^2 * 2 ) * 2 * j^2
<=> n = 3*2*2 * j
<=> n = 12 * j
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 [#permalink] New post 09 Jan 2007, 17:37
shuld be D

n^2/72 = k (an int)

n^2 = 3^2*2^3*k--------------(1)

to make n^2 a perfect square

multiply (1) by 2*a value for k such that k is there in the answer choices and divisible by 72 which is 36

So D :-D
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 [#permalink] New post 09 Jan 2007, 17:58
E for me. This one took me a bit of time to figure out.

n^2 is divisible by 72

Prime Factors of 72 are 2^3*3^2

Hence n^2 = 2^2*3^2*2*k where k is any integer such that 2*k is a square
Let 2*k=j
Therefor the largest number that n is divisible by is 2*3*j=6*j i.e. some multiple of 6

Given the choices the bigggest multiple of 6 is 48 (6*8)i.e. 2*k =64

48^2 is divisible by 72. Hence the largest integer that divides n is 48.
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 [#permalink] New post 09 Jan 2007, 18:11
E ..
48^2/72 = 32..
n= 48 and
48 is the largest possible positive integer that divides n .
:)
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Re: PS: Value of n [#permalink] New post 09 Jan 2007, 19:07
Himalayan wrote:
2. If n is a positive integer and n^2 is divisible by 72, then the largest possible positive integer that must divide n is
A) 6
B) 12
C) 24
D) 36
E) 48


Surely, of the choices given, 48 is the largest # that could divide n, but 12 is the one that must divide n.

The wording is a bit tricky, watch out for must/could/cannot/etc.

Last edited by Andr359 on 09 Jan 2007, 20:40, edited 1 time in total.
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 [#permalink] New post 09 Jan 2007, 19:50
My vote is for B. My approach was the same as Fig's.

n should be a multiple of 12. For any value of n, 12 should divide n. The other nos. MAY or MAY not depending on n.

For example if n=36 then n^2 is divisible by 72, n will not be divisible by 48 but will be divisible by 12. The question asks for the largest number that MUST divide n. So 12 should be the answer.
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 [#permalink] New post 10 Jan 2007, 08:17
i vote for E

48=3*2*2*2
72=3*3*2*2*2

48*48 /72= (3*2*2*2*3*2*2*2 ) /(3*3*2*2*2)= 2*2*2
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 [#permalink] New post 10 Jan 2007, 19:30
Fig wrote:
(B) for me :)

72 = 9*8 = (3)^2 * (2)^2 * 2

n^2 = ( (3)^2 * (2)^2 * 2 ) * k^2
=> n = 3*2*sqrt(2)*k : non sense, as n is an integer, it forces k to not be an integer.

We must have an integer j such that :
n^2 = ( (3)^2 * (2)^2 * 2 ) * 2 * j^2
<=> n = 3*2*2 * j
<=> n = 12 * j


Fig, could you please explain the bolded sections in a little more detail.
For the first one, since n^2 is divisible by 72, shouldn't it be written as
n^2 = ( (3)^2 * (2)^2 * 2 ) * k : (where k is any integer)
The second one, I'm totally lost :roll:
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 [#permalink] New post 10 Jan 2007, 20:01
Let me try.

The prime factors of a square should be squares too.
Eg. 6=3*2, 36=3^2*2^2

The same applies for n^2 - its prime factors should be in squares.
Since n^2 is divisible by 72, n^2 =72*x (an integer)

Finding its prime factors n^2 = 3^2*2^3*x =(3^2)*(2^2)*2*x where x should be 2*y*y to make a square ie (3^2)*(2^2)*(2^2)*(y^2)

So n=3*2*2*y
Note: Without splitting x, n should be 3*2*sqrt(2*x)
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 [#permalink] New post 11 Jan 2007, 01:50
If n is a positive integer and n^2 is divisible by 72, then the largest possible positive integer that must divide n is
A) 6
B) 12
C) 24
D) 36
E) 48

n^2 = 72k ie: 2^3*3^2 k , thus k has to be at least in the for m of 2p,p at least = 1 , p is a perfect square

thus n^2 = 144p ie : n= 12 sqrt p , p is perfect square ( unknown ) thus the largest possible devisor of n is 12

my answer is B , I second FIG on that
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 [#permalink] New post 11 Jan 2007, 07:00
n^2=72k=36*2k = 6^2*2k
2k needs to be the square of a number
k=2m^2, where m is a positive integer
therefore
n=12m

Therefore 12 is the largest possible number that must divide n. Other possible numbers that must divide n would be 1,2,3,4,6. Other possible numbers that may divide n could be anything, depend on what m is.
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 [#permalink] New post 11 Jan 2007, 08:48
I missed the fine print. The question is looking for the largest possible positive integer that "must" divide n.

Hence it has to be 12.

While n "can" be divided by 48 if n is sufficiently large, but 48 will not work if n is small eg. 144

144(n^2) is divisible by 72 Therefor n is divisible by 12. Hence 12 i.e. B should indeed be the right answer.

What is the OA?
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 [#permalink] New post 11 Jan 2007, 19:58
HongHu wrote:
n^2=72k=36*2k = 6^2*2k
2k needs to be the square of a number
k=2m^2, where m is a positive integer
therefore
n=12m

Therefore 12 is the largest possible number that must divide n. Other possible numbers that must divide n would be 1,2,3,4,6. Other possible numbers that may divide n could be anything, depend on what m is.


No wonder you scored 780... :-D

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Re: PS: Value of n [#permalink] New post 12 Jan 2007, 17:22
Himalayan wrote:
2. If n is a positive integer and n^2 is divisible by 72, then the largest possible positive integer that must divide n is
A) 6
B) 12
C) 24
D) 36
E) 48


OA I have is B.

but for me the question is not properly worded as it says largest possible. is 12 the largest possible value that divide n? lets see....

(n^2)/72 = k
n^2 = 72k
n = 6 x sqrt(2k)

now, k could be a minimum of 2 or 2x2x2, 2x3x3, 2x4x4, 2x5x5, 2x6x6 or 2(n^2).

If k = 2, n = 12 and n is divisible by 12.
If k = 2x2x2, n = 24 and n is divisible by 24.
If k = 2x3x3, n = 36 and n is divisible by 36.
If k = 2x4x4, n = 48 and n is divisible by 48.
If k = 2x5x5, n = 60, and n is divisible by 60.

so we know that, n is a multiple of 12 and it is divisible by any number that is a multiple of 12.

where am I missing?????? suggest me.....
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Re: PS: Value of n [#permalink] New post 12 Jan 2007, 18:10
Himalayan wrote:
If n is a positive integer and n^2 is divisible by 72, then the largest possible positive integer that must divide n is
A) 6
B) 12
C) 24
D) 36
E) 48


The problem says the largest possible that must divide n. Let´s solve step by step.
1) n^2 is multiple of 72 => n^2 = 72 * a (a = integer>=1)
2) n^2 = 2^3 * 3^2 * a = 2^2 * 3^2 * 2a
3) n = 2 * 3 * sqrt(2a)
4) n is integer => sqrt(2a) must be integer too. Let´s say that sqrt(2a) = 2*k, k an integer. sqrt(2a) has to be even because an integer sqrt of an even number is always even as well (think 16 and 4, 36 and 6, 144 and 12, etc).
5) n = 2 * 3 * 2k = 4 * 3 * k = 12 * k. k could be 1, 5, 24, 8239823698298, any integer.
6) What is the largest +ve integer that must divide n?
7) Would it be 48? What if k = 5? n = 12 * 5 = 60; can 48 divide 60? No.
8) Which numbers must divide n = 12 * k? 2, 3, 4, 6, and 12.
The trick is to focus 1st on the "must", and only afterwards on the "largest".
Re: PS: Value of n   [#permalink] 12 Jan 2007, 18:10
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