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

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... ???

ScottTargetTestPrep Bunuel i have understood why and how 12 is the answer, but why are we assuming that in order to find largest integer that must divide n, we first need to find the least/minimum possible value of n and not the maximum possible value of n ???

if i take maximum possible value of n ie 48, then 48 will be the largest positive integer that will divide n as per the same logic..

PLEASE HELP me understand this !

Look forward to hearing from you


ScottTargetTestPrep

---------------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

I'm confused - why can't you throw an additional "2" to each so that n = 24 or any of the other answer choices. The question asks for the LARGEST positive integer after all.

"Why are we taking the smallest number and calling it largest".

SOLUTION

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 largest positive integer that must divide \(n\), means for the least value of \(n\) which satisfies the given statement in the question. The lowest square of an integer, which is multiple of \(72\) is \(144\) --> \(n^2=144=12^2=72*2\) --> \(n_{min}=12\). Largest factor of \(12\) is \(12\).

OR:

Given: \(72k=n^2\), where \(k\) is an integer \(\geq1\) (as \(n\) is positive).

\(72k=n^2\) --> \(n=6\sqrt{2k}\), as \(n\) is an integer \(\sqrt{2k}\), also must be an integer. The lowest value of \(k\), for which \(\sqrt{2k}\) is an integer is when \(k=2\) --> \(\sqrt{2k}=\sqrt{4}=2\) --> \(n=6\sqrt{2k}=6*2=12\)

Answer: B.

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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

Given: \(72k=n^2\), where \(k\) is an integer \(\geq1\) (as \(n\) is positive).

\(72k=n^2\) --> \(n=6\sqrt{2k}\), as \(n\) is an integer \(\sqrt{2k}\), also must be an integer. The lowest value of \(k\), for which \(\sqrt{2k}\) is an integer is when \(k=2\) --> \(\sqrt{2k}=\sqrt{4}=2\) --> \(n=6\sqrt{2k}=6*2=12\)

Answer: B.


chetan2u

in 2nd solution ,Bunuel has choosen lowest value of k ...why so?? k can be 8(k=8)

Please help me understand logic

thx

SOLUTION

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 largest positive integer that must divide \(n\), means for the least value of \(n\) which satisfies the given statement in the question. The lowest square of an integer, which is multiple of \(72\) is \(144\) --> \(n^2=144=12^2=72*2\) --> \(n_{min}=12\). Largest factor of \(12\) is \(12\).

OR:

Given: \(72k=n^2\), where \(k\) is an integer \(\geq1\) (as \(n\) is positive).

\(72k=n^2\) --> \(n=6\sqrt{2k}\), as \(n\) is an integer \(\sqrt{2k}\), also must be an integer. The lowest value of \(k\), for which \(\sqrt{2k}\) is an integer is when \(k=2\) --> \(\sqrt{2k}=\sqrt{4}=2\) --> \(n=6\sqrt{2k}=6*2=12\)

Answer: B.

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Hi Bunuel,

what about n=24, n^2=576 which too is divisible by 72. hence in this case the largest integer should be 24. In this question certain limitations should be set which isn't. Sometimes OG questions seems to be off the mark which shouldn't be.

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

SOLUTION

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 largest positive integer that must divide \(n\), means for the least value of \(n\) which satisfies the given statement in the question. The lowest square of an integer, which is multiple of \(72\) is \(144\) --> \(n^2=144=12^2=72*2\) --> \(n_{min}=12\). Largest factor of \(12\) is \(12\).

OR:

Given: \(72k=n^2\), where \(k\) is an integer \(\geq1\) (as \(n\) is positive).

\(72k=n^2\) --> \(n=6\sqrt{2k}\), as \(n\) is an integer \(\sqrt{2k}\), also must be an integer. The lowest value of \(k\), for which \(\sqrt{2k}\) is an integer is when \(k=2\) --> \(\sqrt{2k}=\sqrt{4}=2\) --> \(n=6\sqrt{2k}=6*2=12\)

Answer: B.

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