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If n is a positive integer and n-squared is divisible by 72,

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If n is a positive integer and n-squared is divisible by 72, [#permalink]

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09 Nov 2007, 21:57
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If n is a positive integer and n-squared is divisible by 72, then the largest positive integer that must divide n is
(A) 6
(B) 12
(C) 24
(D) 36
(E) 48

VP
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09 Nov 2007, 22:29
We have to work with n and with 72 where 72 = 2*3*2*2*3.

since (n*n)/72 = integer

then (n*n)/2*3*2*2*3 = integer

n can be 2*3*2 since (2*3*2)*(2*3*2)/(2*3*2*2*3) = 2 = integer

so our answer (not n) has to be 12.

you could ask why our answer can't be 24 since 72 = 2*3*2*2*3.

and (2*3*2*2*3)*(2*3*2*2*3)/(2*3*2*2*3) = 72

and 72/24 = 3

well it can - but the question asks about the largest positive integer that must divide n:

and if n is 12 the answer is 12 but not 24. But if n=72 then our answer can be 6, 12, 24 or 36.

so the largest positive integer that must divide n is 12.

SVP
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09 Nov 2007, 22:39
N^2 = x * 72 (If N^2 is divisible by 72 then x must be positive integer)
= x * 8 * 9
= x * 2 * 2 * 2 * 3 * 3
= 2 * 2 * 2 * 2 * 3 * 3
= 4^2 * 3^2
= 12 ^2

Quote:
If n is a positive integer and n-squared is divisible by 72

From this clause, N can be 12,24,48

Quote:
then the largest positive integer that must divide n is

From this clause: Largest integer that must divide n is 12

Assume largest integer is 24 --- if n is 12.. this statement p integer must divei n is failed.
Same for 48.

12 satisfies both conditions..

So B is correct
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10 Nov 2007, 06:48
i must be missing something here, as i cant fully get either answer.

I can break 72 down into its prime factors, but then what ?
VP
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10 Nov 2007, 06:54
pmenon wrote:
i must be missing something here, as i cant fully get either answer.

I can break 72 down into its prime factors, but then what ?

try to work with the factors (i.e 2*2*3 = 12 and 12*12 = 144 and 144/72 = 2)

You will find that 12 must be the answer.

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If n is a positive integer and n-squared is divisible by 72,

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