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by kop » Wed Dec 18, 2013 2:24 am
If n is a positive integer and n^2 is divisible by 72, then
the largest positive integer that must divide n is

A 16
B 12
C 24
D 36
E 48

Please explain ?
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by GMATGuruNY » Wed Dec 18, 2013 2:39 am
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
n² must be divisible by 72.
Since n² is the square of an integer, the value of n² must be a perfect square: 1, 4, 9, 16, 25, 36, 64, 81, 100, 121, 144...
In the list above, the smallest value divisible by 72 is 144.
If n² = 144, then n = 12.
This is the MINIMUM value of n.
Since n=12 is not divisible by 24, 36, or 48, eliminate C, D and E.
Since n cannot be any smaller than 12, it must be divisible by answer choice B (12).

The correct answer is B.
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by theCodeToGMAT » Wed Dec 18, 2013 3:12 am
72 = 2x2x2 x 3x3

"n" must have at least = two 2's & one 3 ==> 2x2x3 = 12

So, 12

[spoiler]{B}[/spoiler]
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by madhavanjc » Wed Dec 18, 2013 5:28 am
How can you say that it should have two 2's and one 3...??
can u elaborate.??
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by theCodeToGMAT » Wed Dec 18, 2013 5:59 am
madhavanjc wrote:How can you say that it should have two 2's and one 3...??
can u elaborate.??
We know that "n" is a positive integer and n^2 is divisible by 72

and, 72 = 2x2x2 x 3x3

So, we need minimum three 2's and two 3's in n^2

If "n" were 2x3 .. then n^2 = 2x2 x 3x3 .. which is not divisible by "72".. we are laging with one "2".. NO

but,

If "n" were 2x2x3.. then n^2 = 2x2 x 2x2 x 3x3.... which is divisible by "72" since we got minimum number of 2's & 3's..... YES

Another way to look at this is .. take sqrt of "72" ==> sqrt(2x2x2 x 3x3) ==> 2 x 3 x sqrt(2).... so we must have n as 2x3x2 = 12
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