Properties of numbers question

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Properties of numbers question

by kellogs4toniee » Mon Jan 23, 2012 4:15 pm
There is a question which has an explanation that I am not entirely clear on.

Question :

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


Explanation :

Answer is 12. Since n^2 is divisible by 72, n^2 = 72k for some positive integer k. Since n^2 = 72k, then 72k must be a perfect square. Since 72k = (2^3)(3^2)k, then k = 2m^2 for some positive integer m in order for 72k to be a perfect square. Then, n^2 = 72k = (2^3)(3^2)(2m^2) = (2^4)(3^2)m^2 = ((2^2)(3)(m))^2, and n = (2^2)(3)m. The positive integers that MUST divide n are 1,2,3,4,6 and 12. Therefore, the answer is 12.


The part I get confused on is where it talks about k=2m^2 for some positive integer m in order for 72k to be a perfect square. I am fairly sure there is a simpler or more straight forward explanation and am hoping someone can help me clear this up. Or what kind of mindset should I be going with when solving this kind of problem.

Much appreciated in advance!
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by pemdas » Mon Jan 23, 2012 7:05 pm
I agree with you the explanation offered is quite lengthy and doesn't help to focus on the main takeaways from this question. There are two major takeaways:

1) n^2/72 doesn't imply divisibility for n/Sqrt(72), hence they construe n^2=72K and further on their way build such an explanation indicated above.
2) the question is limiting the possible largest divisors to MUST divide

In the explanation offered 2m^2 is introduced and afterward rejected for one simple reason to make sure we match 'MUST divide' condition for the largest divisor. As such I would suggest to decide that we are missing one 2 here and the rest perfect squares (m) are different possibilities (not MUST). Hence 72*2=144 is a perfect square or the least possible value (72*2) here and sqrt(144)=12.
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by Anurag@Gurome » Mon Jan 23, 2012 7:58 pm
kellogs4toniee wrote:There is a question which has an explanation that I am not entirely clear on.

Question :

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


Explanation :

Answer is 12. Since n^2 is divisible by 72, n^2 = 72k for some positive integer k. Since n^2 = 72k, then 72k must be a perfect square. Since 72k = (2^3)(3^2)k, then k = 2m^2 for some positive integer m in order for 72k to be a perfect square. Then, n^2 = 72k = (2^3)(3^2)(2m^2) = (2^4)(3^2)m^2 = ((2^2)(3)(m))^2, and n = (2^2)(3)m. The positive integers that MUST divide n are 1,2,3,4,6 and 12. Therefore, the answer is 12.


The part I get confused on is where it talks about k=2m^2 for some positive integer m in order for 72k to be a perfect square. I am fairly sure there is a simpler or more straight forward explanation and am hoping someone can help me clear this up. Or what kind of mindset should I be going with when solving this kind of problem.

Much appreciated in advance!
72k = n² for some positive integer k (as n is also a positive integer).
72k = n² implies n = √(72k) = 6√(2k)
Now since n is a positive integer, so √(2k) should also be a integer. The minimum value of k for which √(2k) will be an integer is for k = 2.
So, √(2k) = √4 = 2, which implies n = 6√(2k) = 6 * 2 = 12

The correct answer is B.
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by pemdas » Mon Jan 23, 2012 8:51 pm
the least for value for n is when k=2 for 6√(2k) and n = 6 * 2 = 12
n can be greater than 12 too
Anurag@Gurome wrote: So, √(2k) = √4 = 2, which implies n = 6√(2k) = 6 * 2 = 12

The correct answer is B.
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