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Properties of numbers Quant Review 2nd Ed #169

Expert replies
by runningguy » Fri Sep 20, 2013 1:54 pm
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
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Source: — Problem Solving |

by Brent@GMATPrepNow » Fri Sep 20, 2013 2:19 pm
runningguy 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
The key word here is MUST.
In order to find the largest positive integer that must divide n, we need to find the smallest possible value of n.

The question tells us that n² is divisible by 72.
So, n² is a multiple of 72 AND n² is the square of an integer.
Let's list some squares of integers: 1, 4, 9, 16, . . . 100, 121, 144, 169...
So, 144 is the smallest perfect square that is divisible by 72
If n² = 144, then n = 12.
So, 12 is the smallest possible value of n that satisfies the given conditions.

If n = 12, then the largest positive integer that must divide n is 12 = B

Aside: Notice that 12 is NOT DIVISIBLE by 24, 36 or 48 (so we can eliminate C, D and E). This shows why we wanted to find the smallest possible value of n.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by GMATGuruNY » Fri Sep 20, 2013 2:20 pm
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 » Sat Sep 21, 2013 5:20 am
n^2/72
72= 2x2x3x3x2 = 2^3x3^2

we need 2^2x3.. So, 12

ANSWER [spoiler][/spoiler]
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