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BTG 300 unclear solution

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by replayyyy » Mon Oct 25, 2010 10:21 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

The problem is from BTG 300 tough problems file, #28 and the OA is B - 12. However, I can`t understand why this has to be true because if n=72, for example, then 72 itself divides n and this clearly is bigger than 12 ???
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by fskilnik@GMATH » Mon Oct 25, 2010 10:59 am
replayyyy 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

I can`t understand why this has to be true because if n=72, for example, then 72 itself divides n and this clearly is bigger than 12 ???
Hi, replayyyy!

Good argument of yours, really, but your mistake is in the question stem interpretation! If it is looking for the largest positive integer that MUST divide n, if n is for instance equal to 12, then is is true that "n is a positive integer and n^2 is divisible by 72" , but it is not true that 72 divides 12 ...

In other words, we are looking for a solution (alternative choice) that is guaranteed for all n that satisfies the question stem... understood?

In the following post I will offer a possible solution to the problem, because this one was "just" to tell you that your argument was not able to "invalidate the exercise"... ok?!

Regards,
Fabio.
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by Brian@VeritasPrep » Mon Oct 25, 2010 11:03 am
Hey replayyy,

I love questions like this - thanks for sharing it!

If n^2 is divisible by 72 and n is an integer, then let's calculate the potential values of n, which we'd get by taking the root of 72:

sqrt 72 = 6 * sqrt 2

Well, the square root of 2 is not an integer, so in order for n to be an integer that term must be a 2. So we can prove that 6 * 2 is a factor of n.

You can also look at it this way - the factors of n^2 are the factors of 72: 2*2*2*3*3

When we take the square root of n^2 to determine the factors of n (n^2 = n*n, so we need to break off half the factors to divide them into individual ns), we're left we can split the pairs 2*2 and 3*3, but the remaining 2 can't be split into integers for each individual n. Therefore there must be another 2 to allow for n to be an integer.

That's why this statement n^2 is divisible by 72 means that integer n must be divisible by 12. The other factor of 2 is implied by the statement "n is an integer" because that's the only way to make it correct. Tricky problem, but definitely the kind of thing that they'll love to ask.
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by fskilnik@GMATH » Mon Oct 25, 2010 11:08 am
replayyyy 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
72 equals 2.36, that is, (2^3)(3^2) , therefore if n is a positive integer such that its square is a multiple of 72, that means that n^2 is (2^3)(3^2).(positive integer M), where M is such that:

> factor 2 appears in M and it is in an odd number of times (equal or greater to 1)
> if factor 3 appears in M, it appears in an even number of times (equal or greater to 2)

(This is due to the fact that a perfect square has each of its prime factors always presented in an even number of times.)

Now we are looking for the GREATEST divisor GUARANTEED of n, that means we should imagine (and IMPOSE) that n has the LEAST it has to have... (think about it), therefore let us impose that M has just one factor 2, no factor 3, and nothing else!!!

Finally n^2 is (2^3)(3^2).2 = (2^4)(3^2) implying (n is positive) that n = (2^2).(3^1), that is, n is 12 (minimum) so that the greater possible divisor of n is really 12... the answer is correct.

I hope you like it. Beautiful problem, by the way.

Regards,
Fabio.
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by GMATGuruNY » Mon Oct 25, 2010 11:17 am
replayyyy 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 problem is from BTG 300 tough problems file, #28 and the OA is B - 12. However, I can`t understand why this has to be true because if n=72, for example, then 72 itself divides n and this clearly is bigger than 12 ???
The trick to solving most divisibility questions is prime-factorization.

72 = 2^3 * 3^2
Thus, for n^2 to be divisible by 72, it must be divisible by 2^3 and by 3^2.
The smallest integer value that will work for n is 2*2*3, because then n^2 = (2*2*3)^2 = 2^4 * 3^2, which will be divisible by 2^3 and by 3^2. (If n = 2*3, n^2 = (2*3)^2 = 2^2 * 3^2, which will not be divisible by 2^3.)

If n = 2*2*3 = 12 (the smallest value it could be), then it is not divisible by 24, 36, or 48. Eliminate C, D and E. The largest positive integer that must be a factor of n=12 is 12.

The correct answer is B.
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by replayyyy » Mon Oct 25, 2010 12:34 pm
Thanks! Comprehensive explanations!
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by jeetu_vishnoi » Mon Oct 25, 2010 2:31 pm
if we factorise 72 then it is 2*2*2*3*3
it is given that n2 is divisible by 72 hence if we make the above facorisation as square number then we have to multiply it by 2:
(2*2)(2*2)(3*3)
now from here you can see it easily that the largest number which can divide n is 2*2*3 = 12
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by blue.vikasbhardwaj@gmail. » Wed Oct 27, 2010 5:03 am
replayyyy 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 problem is from BTG 300 tough problems file, #28 and the OA is B - 12. However, I can`t understand why this has to be true because if n=72, for example, then 72 itself divides n and this clearly is bigger than 12 ???
If n^2 is divisible by 72 , then

n^2 = 72k , where k =1,2,3....

n = (72k)^1/2

n = 6 * (2k)^1/2

Since n is a positive integer , then (2k)^1/2 must be an perfect square

So , let k = 2 * m^2 , where m = 1,2,3.....

Therefore , n becomes 12m , where m =1,2,3....

Thus largest number with which n is always divided is 12.

Ans B
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