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If n is a positive integer and n^2 is divisible by 72, then

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Re: If n is a positive integer and n^2 is divisible by 72, then  [#permalink]

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New post 16 Nov 2017, 15:13
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amitvmane 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

---------------ASIDE #1--------------------------------------
A lot of integer property questions can be solved using prime factorization.
For questions involving divisibility, divisors, factors and multiples, we can say:

If N is a factor by k, then k is "hiding" within the prime factorization of N

Consider these examples:
3 is a factor of 24, because 24 = (2)(2)(2)(3), and we can clearly see the 3 hiding in the prime factorization.
Likewise, 5 is a factor of 70 because 70 = (2)(5)(7)
And 8 is a factor of 112 because 112 = (2)(2)(2)(2)(7)
And 15 is a factor of 630 because 630 = (2)(3)(3)(5)(7)

---------------ASIDE #2--------------------------------------
IMPORTANT CONCEPT: The prime factorization of a perfect square will have an even number of each prime

For example: 400 is a perfect square.
400 = 2x2x2x2x5x5. Here, we have four 2's and two 5's
This should make sense, because the even number of primes allows us to split the primes into two EQUAL groups to demonstrate that the number is a square.
For example: 400 = 2x2x2x2x5x5 = (2x2x5)(2x2x5) = (2x2x5)²

Likewise, 576 is a perfect square.
576 = 2x2x2x2x2x2x3x3 = (2x2x2x3)(2x2x2x3) = (2x2x2x3)²
--------NOW ONTO THE QUESTION!------------------

Given: n² is divisible by 72 (in other words, there's a 72 hiding in the prime factorization of n²)
So, n² = (2)(2)(2)(3)(3)(?)(?)(?)(?)(?)... [the ?'s represent other possible primes in the prime factorization of n²]
Since we have an ODD number of 2's in the prime factorization, we can be certain that there is at least one more 2 in the prime factorization.
So, we know that n² = (2)(2)(2)(3)(3)(2)(?)(?)(?)(?)
So, while there MIGHT be tons of other values in the above prime factorization, we do know that there MUST BE at least four 2's and two 3's.
Now do some grouping to get: n² = [(2)(2)(3)(?)(?)...][(2)(2)(3)(?)(?)...]
From this we can see that n = (2)(2)(3)(?)(?)...

Question: What is the largest positive integer that must divide n?
(2)(2)(3) = 12.
So, the largest positive integer that must divide n is 12

Cheers,
Brent
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Re: If n is a positive integer and n^2 is divisible by 72, then  [#permalink]

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New post 14 Dec 2017, 20:42
YYZ wrote:
ScottTargetTestPrep wrote:
amitvmane 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


We are given that n^2/72 = integer or (n^2)/(2^3)(3^2) = integer.

However, since n^2 is a perfect square, we need to make 72 or (2^3)(3^2) a perfect square. Since all perfect squares consist of unique primes, each raised to an even exponent, the smallest perfect square that divides into n^2 is (2^4)(3^2) = 144.

Since n^2/144 = integer, then n/12 = integer, and thus the largest positive integer that must divide n is 12.

Answer: B


Hey Scott,

I think we may have covered it buy why can't we make n=72 therefore n^2 = (72)(72) ... why are we trying to use the SMALLEST perfect square .. using n=72 follows the rules set out in the question... ???


The question asks to find the largest positive integer that MUST divide n. So, which ALWAYS divides n, if n^2 is divisible by 72. Now, while n COULD be divisible by any integer, for example, by by 48, 72, 1,000,000, ... it MUST be divisible only by factors of 12. Why? Because the least value of n for which n^2 is divisible by 144 is 12.

Hope it's clear.
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If n is a positive integer and n^2 is divisible by 72, then  [#permalink]

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New post 15 Dec 2017, 16:18
ScottTargetTestPrep wrote:

We are given that n^2/72 = integer or (n^2)/(2^3)(3^2) = integer.

However, since n^2 is a perfect square, we need to make 72 or (2^3)(3^2) a perfect square. Since all perfect squares consist of unique primes, each raised to an even exponent, the smallest perfect square that divides into n^2 is (2^4)(3^2) = 144.

Since n^2/144 = integer, then n/12 = integer, and thus the largest positive integer that must divide n is 12.

Answer: B




Got it!!! Thank you!
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Re: If n is a positive integer and n^2 is divisible by 72, then  [#permalink]

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New post 21 Jun 2018, 07:30
ziko wrote:
Normally i divide the number into the primes just to see how many more primes we need to satisfy the condition, so in our case:
n^2/72=n*n/2^3*3^2, in order to have minimum in denominator we should try modify the smallest number. If we have one more 2 then the n*n will perfectly be devisible to 2^4*3^2 from here we see that the largest number is 2*2*3=12
Hope i explained my thought.



Had the question been n^3/72, then the answer would have been 3 right?
Re: If n is a positive integer and n^2 is divisible by 72, then &nbs [#permalink] 21 Jun 2018, 07:30

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