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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.

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.

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.

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