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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 01 Nov 2006, 12:48
If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is?

a) 6
b)12
c)24
d)36
e)48

I can't seem to fully understand the OE.
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 [#permalink] New post 01 Nov 2006, 13:05
The largest possible integer that divides a given integer is the integer itself.
Now out of the given values of n we need to larget value of n such n^2 is divisible by n. The condition holds true for B,C,D,E and largest among them is E. So my pick is E.

What is OA?
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 [#permalink] New post 01 Nov 2006, 13:05
If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is?

a) 6 b)12 c)24 d)36 e)48

n^2 = 2^3*3^2 k ( K MUST AT LEAST BE IN THE FORM 2X)

then the largest is 12 my answer is B
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 [#permalink] New post 01 Nov 2006, 13:07
yezz wrote:
If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is?

a) 6 b)12 c)24 d)36 e)48

n^2 = 2^3*3^2 k ( K MUST AT LEAST BE IN THE FORM 2X)

then the largest is 12 my answer is B

yezz

48 =12*4 and 48^2 =12*12*4*4 which will be divisble by 72.
So why not E then?
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 [#permalink] New post 01 Nov 2006, 13:14
If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is?

a) 6 b)12 c)24 d)36 e)48

N = +VE SQRT OF N^2

N^2= 72K = 2^3*3^2*K AND SINCE N IS INTIGER THEN

N^2 MUST AT LEAST = 144B WHERE B IS +VE PERFECT SQURE INTIGER

THUS N IS AT LEAST= 12G

THE KEY IS ( MUST DEVIDE) , WE ONLY KNOW FROM ABOVE THAT 12 IS

BECAUSE G COULD BE ANYTHING.

Hope this helps
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 [#permalink] New post 01 Nov 2006, 13:17
Yezz you are right...
Forgot the must in question. :oops:
Thanks for the explanation

:-D
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 [#permalink] New post 01 Nov 2006, 14:20
yezz wrote:
If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is?

a) 6 b)12 c)24 d)36 e)48

N = +VE SQRT OF N^2

N^2= 72K = 2^3*3^2*K AND SINCE N IS INTIGER THEN

N^2 MUST AT LEAST = 144B WHERE B IS +VE PERFECT SQURE INTIGER

THUS N IS AT LEAST= 12G

THE KEY IS ( MUST DEVIDE) , WE ONLY KNOW FROM ABOVE THAT 12 IS

BECAUSE G COULD BE ANYTHING.

Hope this helps


OA is B, but I am still having a hard time understanding

I think I misunderstood the question . . . so basically, what you're trying to do is find the smallest multiple of 72 that is a perfect square. That way you figure out the largest divisor that (MUST) always divides into n.

Like Yogethsheth, I was thinking largest possible.
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 [#permalink] New post 02 Nov 2006, 02:58
N= 12 G THUS IT CAN GIVE INFINITE ( LARGEST POSSIBLE)
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 [#permalink] New post 02 Nov 2006, 08:17
I worked this problem using an example, 144 is the least perfect square divisible by 72, and 12 is the square root of 144. So, 12 is the largest integer that can perfectly divide 12.

Therefore, my answer is B.[/code]
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 [#permalink] New post 02 Nov 2006, 12:13
nischalb wrote:
I worked this problem using an example, 144 is the least perfect square divisible by 72, and 12 is the square root of 144. So, 12 is the largest integer that can perfectly divide 12.

Therefore, my answer is B.[/code]


exactly the logic i used..

if n2 is divisible by 72 then N^2 is divisible by multiples of 72.

The smallest possible perfect sqr is 144 which is divisible by 72

Hence the ans 12
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 [#permalink] New post 03 Nov 2006, 22:41
the key here is MUST, 48 may but 12 must.

n^2 / 2^3 * 2^2 , so at a minimum n = 2^2 * 3 as it is the smallest number when squared that is divisible by 72

so 12
  [#permalink] 03 Nov 2006, 22:41
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