Number properties question

yes, is there a rule? what if it was n^3 or n^4 - I can see that you can work through the question since it's quite straight forward but i would have no idea how to do this if the numbers were a lot more complex

n^2 = 96(..)

= (2 ^5)(3)(...)
to make it a perfect square there has to to atleast one more 2 and 3 in n^2.

=> n will be atleast 2^3*3 = 24

so largest positive integer that must divide n is 24.

Lets assume n to be 2^2 * 3 i.e. 12
Now n^2
=> 12^2/96
=> 144/96
remainder is 48. for n to be divisible by 96, remainder should be zero.

same method can prove that 2*3 doesn't satisfy. On the contrary, the question can be why not a large value like 2^10*3^2 - this part would depend on the answer choices listed, without list of ans choices, we can solve for a min value.

Alternate method, just thought of it - may not be the best way.

=>n^2 / 96
=> n^2 / (2^5 x 3^1)
=> n^2 x 2^(-5) x 3 ^ (-1) ---- call it eq A

For the above to be divisible (leave remainder=zero), there should be no negative powers left after addition/subtraction of powers

1) Let n be 2^2 x 3
n^2 = (2^2 x 3)^2
=> 2^4 x 3^2
Substituting above in eq A for n^2
=>2^4 x 3^2 x 2^(-5) x 3 ^ (-1)
=>2^(4 - 5) x 3^(2 - 1)
=> 2^(-1) x 3^1 or 3/2
so the assumed value of n is incorrect, we can increase the number of 2's in n so that denominator is 1

Another method that crossed my mind, may look silly though -
Taking log of n^2 / (2^5 x 3^1), the result should be >=0
=>2log(n) - 5log2 - log3
substituting n with (2^3 x 3)^2 i.e. 2^6 x 3^2
=>6log2 + 2log3 - 5log2 - log3
=> log2 + log3
or log (2 x 3)
removing log => 6
positive value indicates it is

For reference,
log of a^b = b log a
log of a x b = log a + log b
log of a/b = log a - log b

\(n^2\) is divisible by \(96\) i.e. \(2^5*3\)

What is minimum value for \(n^2\) i.e. which is the smallest number must be multiplied by \(2^5*3\) to make it a perfect square?

We know the powers of all prime numbers must be even in a perfect square.
\(2^5\) if multiplied by \(2\) will result in even power of 2 i.e. \(2^6\)
\(3^1\) if multiplied by \(3\) will result in even power of 3 i.e. \(3^2\)

Thus, we multiplied \(2^5*3\) by \(2*3\). n^2 becomes \(2^6*3^2\), which a square of \(2^3*3\)

\((2^3*3)^2=2^6*3^2\)

Thus, we have our minimum value for n.
\(2^3*3=24\)

The largest integer that divides 24 is 24 itself. Smaller integers would be 2,12,4,6,1. Ans: 24(Supposedly)

Now, to add a little more to the confusion.

It is possible that \(n^2=((2^3*3)^2)^2 \hspace{} OR \hspace{2} ((2^3*3)^{25})^2\)
Making \(n=(2^3*3)^2 \hspace{} OR \hspace{2} (2^3*3)^{25} \hspace{2}\)
Eventually making largest integer that could divide \(n=(2^3*3)^2 OR \hspace{2} (2^3*3)^{25}\)

We don't know the exact value of \(n\). Thus, we need to consider it as minimum possible because the question asks the largest positive integer that MUST divide n.

Imagine we answered it as \((2^3*3)^{25}\) but in reality, \(n=(2^3*3)\), our answer would be wrong because \((2^3*3)^{25}\) doesn't divide \((2^3*3)\).

At the same time, if we answered it as \((2^3*3)\), it will divide "n", alright. But, even here we can't be too sure whether this is indeed the LARGEST integer that is dividing "n". What if \(n=(2^3*3)^{25}\).

I would have liked the question if it read:
"If n is a positive integer and n^2 is the smallest integer divisible by 96, then the largest positive integer that must divide n will be?

@subhashghosh - As per your request, I am providing the explanation to this question though I think fluke has already done a good job above. I don't think there is much left to add.

Anyway, since \(n^2\) is a multiple of 96,
\(n^2 = 96*k = 2^5 * 3 * k\) (k is a positive integer)

Notice that it is \(n^2\) i.e. square of a positive integer so powers of all prime factors of \(n^2\) must be even. So k must be at least 2*3 so that
\(n^2\) is at least \(2^6 * 3^2\) (note that powers of 2 and 3 have become even now. Also, this is the smallest value of \(n^2. If k > 2*3, n^2\) would be greater)

(If this concept is unclear, check: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2010/12 ... t-squares/)

Since \(n^2\) is at least \(2^6 * 3^2\), \(n\) is at least \(2^3 * 3\). So largest number that must divide n is 24. Note that a greater number could divide n (depending on the value of k)

Hi Fluke,

Thanks a lot for your explanation, really helped.
However, how did you find the ^25 as the other possible value of n^2?

Thanks

\(n^2\) can't be \((2^3*3)^{25}\)

\(n^2\) should be any even power of \((2^3*3)\), say \((2^3*3)^{2k}\) where k is a positive integer. Because, \(n^2\) is a perfect square and a perfect square is always an even power of an integer.

\(2^2=4\)
\(3^2=9\)
\((2^2)^2=16\)
\((3^2)^2=81\)

Thus, \(n^2\) can be \(((2^3*3)^{25})^2\).
If \(n^2=((2^3*3)^{25})^2\), then \(n=\sqrt{((2^3*3)^{25})^2}=(2^3*3)^{25}\)

Now,
Question says; \(n^2\) is divisible by 96.
\(n=(2^3*3)^{25}\)
\(n^2=((2^3*3)^{25})^2\).
\(((2^3*3)^{25})^2\) is divisible by 96 because \((2^3*3)^{2}\) is divisible by 96.

What will be the largest integer that will divide "n". The number itself, right.
Largest integer\(=n=(2^3*3)^{25}\)

Well, this answer is not correct though because the question assumes the smallest value for \(n^2=(2^3*3)^2\), which makes the largest \(n=2^3*3\) I don't really understand, why so?

But, hey!!! This question is from GMATPrep and it is a general belief that GMATPrep questions are always logically correct. You may as well ignore all this gibberish.

Focus on the difference between 'must' and 'can'

Say, if I tell you that a = 20k (k is a positive integer)
My question is 'Which is the largest number by which 'a' must be divisible?'
From this information, you can say that 'a' must be divisible by 20. It could be divisible by 40, 80, 120, 808080 etc