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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
A. 6
B. 12
C. 24
D. 36
E. 48
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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
We are asked to find the largest positive integer that must divide \(n\). For this divisor to apply to all possible values of \(n\), it must divide the smallest possible \(n\) that satisfies the condition. In other words, we need to determine the least value of \(n\) for which \(n^2\) is divisible by 72.
The smallest square of an integer that is a multiple of 72 is 144:
\(n^2 = 144 = 12^2 = 72 * 2\)
Therefore, \(n_{min} = 12\), and the largest factor of 12 is 12.
OR:
Given: \(72k = n^2\), where \(k\) is an integer \(\geq 1\) (as \(n\) is positive).
From \(72k = n^2\), we get \(n = 6\sqrt{2k}\). For \(n\) to be an integer, \(\sqrt{2k}\) must also be an integer. The lowest value of \(k\) that makes \(\sqrt{2k}\) an integer is when \(k = 2\), which gives:
\(\sqrt{2k} = \sqrt{4} = 2\)
Therefore, \(n = 6\sqrt{2k} = 6 * 2 = 12\).
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
We are asked to find the largest positive integer that must divide \(n\). For this divisor to apply to all possible values of \(n\), it must divide the smallest possible \(n\) that satisfies the condition. In other words, we need to determine the least value of \(n\) for which \(n^2\) is divisible by 72.
The smallest square of an integer that is a multiple of 72 is 144:
\(n^2 = 144 = 12^2 = 72 * 2\)
Therefore, \(n_{min} = 12\), and the largest factor of 12 is 12.
OR:
Given: \(72k = n^2\), where \(k\) is an integer \(\geq 1\) (as \(n\) is positive).
From \(72k = n^2\), we get \(n = 6\sqrt{2k}\). For \(n\) to be an integer, \(\sqrt{2k}\) must also be an integer. The lowest value of \(k\) that makes \(\sqrt{2k}\) an integer is when \(k = 2\), which gives:
\(\sqrt{2k} = \sqrt{4} = 2\)
Therefore, \(n = 6\sqrt{2k} = 6 * 2 = 12\).
Answer: B.
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amitvmane
---------------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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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.
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.
catennacio
Only restriction we have on positive integer n is that n^2 is divisible by 72. The least value of n for which n^2 is divisible by 72 is 12, thus n must be divisible by 12 (n is in any case divisible by 12). For all other values of n, for which n^2 is divisible by 72, n will still be divisible by 12. This means that n is always divisible by 12 if n^2 is divisible by 72.
Now, ask yourself: if n=12, is n divisible by 48? No. So, n is not always divisible by 48.
Hope it's clear.
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I approach this problem by prime factorisation.
any square must have 2 pairs of prime factors.
prime factorisation of 72 has 2*2, 3*3 and 2. n^2 must have one more 2 as a prime factor. Hence lasrgest number which must devide n is 2*3*2 = 12
any square must have 2 pairs of prime factors.
prime factorisation of 72 has 2*2, 3*3 and 2. n^2 must have one more 2 as a prime factor. Hence lasrgest number which must devide n is 2*3*2 = 12
Merging similar topics.
The question asks about "the largest positive integer that MUST divide n", not COULD divide n. Since the least value of n for which n^2 is a multiple of 72 is 12 then the largest positive integer that MUST divide n is 12.
Complete solution of this question is given above. Please ask if anything remains unclear.
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raviram80
The question asks about "the largest positive integer that MUST divide n", not COULD divide n. Since the least value of n for which n^2 is a multiple of 72 is 12 then the largest positive integer that MUST divide n is 12.
Complete solution of this question is given above. Please ask if anything remains unclear.
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General Discussion
Hi All
I have a confusion about this question
169. If n is a positive integer and n2 is divisible by 72, then
the largest positive integerthat must divide n is
(A) 6
(8) 12
(C) 24
(0) 36
(E) 48
If we are looking for largest positive integer that must divide n, why can it not be 48.
because if n2 = 72 * 32 then n will be 48 , so does this not mean n is divisible by 48.
Please explain.
Thanks
I have a confusion about this question
169. If n is a positive integer and n2 is divisible by 72, then
the largest positive integerthat must divide n is
(A) 6
(8) 12
(C) 24
(0) 36
(E) 48
If we are looking for largest positive integer that must divide n, why can it not be 48.
because if n2 = 72 * 32 then n will be 48 , so does this not mean n is divisible by 48.
Please explain.
Thanks
Posts: 34
Bunuel
I spent a few hours on this one alone and I'm still not clear. I chose 12 at first, but then changed to 48.
I'm not a native speaker, so here is how I interpreted this question: "the largest positive integer that must divide n" = "the largest positive factor of n". Since n is a variable (i.e. n is moving), so is its largest factor. Please correct if I'm wrong here.
I know that if n = 12, n^2 = 144 = 2 * 72 (satisfy the condition). When n = 12, the largest factor of n is n itself, which is 12. Check: 12 is the largest positive number that must divide 12 --> true
However if n = 48, n^2 = 48 * 48 = 32 * 72 (satisfy the condition too). When n = 48, the largest factor of n is n itself, which is 48. Check: 48 is the largest positive number that must divide 48 --> true
So, I also notice that the keyword is "MUST", not "COULD". The question is, why is 48 not "MUST divide 48", but instead only "COULD divide 48"? I'm not clear right here. Why is 12 "MUST divide 12"? What's the difference between them?
Thanks,
Caten
Posts: 34
Bunuel
Thank you very much Bunuel. Very clear now. Now I understand what "must" means. It means it will be always true regardless of n. As you said (and I chose), when n = 24 or 36 or 48, the answer 48 can divide 48, but cannot divide 12. So the "must" here is not maintained. In this case we have to choose the largest factor of the least possible value of n to ensure that largest factor is also a factor of other values of n. Therefore the least value of n is 12, the largest factor of 12 is also 12. This factor also divides other n values, for all n such that n^2 = 72k.
My mistake was that I didn't understand the "must" wording and didn't check whether my answer 48 can divide ALL possible values of n, including n=12. This is what "must" mean.
Again, thanks so much!
Caten
catennacio
More must/could be true questions from our question banks (viewforumtags.php) here: search.php?search_id=tag&tag_id=193
Hope it helps.
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The explanations above are much clearer to me than the explanation offered in the official Gmat guide;
"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."
Why must k=2m^2?
"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."
Why must k=2m^2?
actionj
A perfect square has its primes in even powers, thus k must complete odd power of 2 into even, hence 2, and it also can have some other integer in even power, hence m^2.
Hope it's clear.
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actionj
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\(\frac{n*n}{72}\) = \(\frac{n*n}{2*2*2*3*3}\)
This means n should be at least 2*2*3. Otherwise \(n^2\) won't be divisible by 72
What is the largest integer that can divide n?
Let's see, largest integer that can divide 2? 2
Largest integer that can divide 3? 3
Largest integer that can divide n is n itself
Answer B
Time Taken 2:31
Difficulty level 650
This means n should be at least 2*2*3. Otherwise \(n^2\) won't be divisible by 72
What is the largest integer that can divide n?
Let's see, largest integer that can divide 2? 2
Largest integer that can divide 3? 3
Largest integer that can divide n is n itself
Answer B
Time Taken 2:31
Difficulty level 650
arunspanda
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Updated on: 23 Mar 2014, 19:01
Originally posted by arunspanda on 18 Mar 2014, 07:00.
Last edited by arunspanda on 23 Mar 2014, 19:01, edited 2 times in total.
Last edited by arunspanda on 23 Mar 2014, 19:01, edited 2 times in total.
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 that, n is a positive integer and n^2 is divisible by 72 means that n^2 is a perfect square that is lowest multiple of 72.
Or, the lowest multiple of 72 that is a perfect square is 144.
So, n^2 = 144q, q is some positive integer => 12 is a factor of n.
Answer: (B)
(A) 6
(B) 12
(C) 24
(D) 36
(E) 48
Given that, n is a positive integer and n^2 is divisible by 72 means that n^2 is a perfect square that is lowest multiple of 72.
Or, the lowest multiple of 72 that is a perfect square is 144.
So, n^2 = 144q, q is some positive integer => 12 is a factor of n.
Answer: (B)
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Bunuel
Hi,
How can the part highlighted in bold be concluded fro the question stem "The largest positive integer that must divide n, means for the least value of n"??
In absence of this even 48 looks like a valid answer, as 48^2 is divisible by 72 and 48 is divisible by itself.
Please help!!
Regards,
Sagar
itsworththepain
We need the largest positive integer that MUST divide n. So, we should find the least value of n for which n^2 is divisible by 72, and if that n is divisible by some number then so will be every other n's (MUST condition will be satisfied). The least positive n for which n^2 is divisible by 72 is 12 (12^2 = 144, which is divisible by 72). So, even if n = 12 it is divisible by 12 but not divisible by 24, 36 or 48.
Follow the links in my previous post for similar questions to understand the concept better.
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In questions like this,
It's a good idea to start with the prime factorized form of n.
We can write \(n = p1^m*p2^n*p3^r\). . . where p1, p2, p3 . . .are prime factors of n and m, n, r are non-negative integers (can be equal to 0)
So, \(n^2 = p1^{2m}*p2^{2n}*p3^{2r}\). . .
Now, \(n^2\) is completely divisible by 72 = \(2^3*3^2\)
This means, \(\frac{(p1^{2m}*p2^{2n}*p3^{2r} . . . )}{(2^3*3^2)}\) is an integer.
What does this tell you?
That p1 = 2 and 2m ≥ 3, that is m ≥ 3/2. But m is an integer. So, minimum possible value of m =2
Also, p2 = 3 and 2n ≥ 2. That is, n ≥ 1. So, minimum possible value of n = 1
Let's now apply this information on the expression for n:
n = \(2^2*3^1\)\(*something. . .\)
From this expression, it's clear that n MUST BE divisible by \(2^2*3^1\) = 12.
Takeaway: If you find yourself getting confused in questions that gives divisibility information about different powers of a number, start by writing a general prime factorized expression for the number raised to power 1.
Hope this was useful!
Japinder
_________________
It's a good idea to start with the prime factorized form of n.
We can write \(n = p1^m*p2^n*p3^r\). . . where p1, p2, p3 . . .are prime factors of n and m, n, r are non-negative integers (can be equal to 0)
So, \(n^2 = p1^{2m}*p2^{2n}*p3^{2r}\). . .
Now, \(n^2\) is completely divisible by 72 = \(2^3*3^2\)
This means, \(\frac{(p1^{2m}*p2^{2n}*p3^{2r} . . . )}{(2^3*3^2)}\) is an integer.
What does this tell you?
That p1 = 2 and 2m ≥ 3, that is m ≥ 3/2. But m is an integer. So, minimum possible value of m =2
Also, p2 = 3 and 2n ≥ 2. That is, n ≥ 1. So, minimum possible value of n = 1
Let's now apply this information on the expression for n:
n = \(2^2*3^1\)\(*something. . .\)
From this expression, it's clear that n MUST BE divisible by \(2^2*3^1\) = 12.
Takeaway: If you find yourself getting confused in questions that gives divisibility information about different powers of a number, start by writing a general prime factorized expression for the number raised to power 1.
Hope this was useful!
Japinder
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Bunuel
CONCEPT: Perfect Squares always have even power of Prime no. in their Prime factorised form.
\(72 = 8*9 = 2^3*3^2\)
Since power of prime numbers must be prime so \(72\)must be multiplied by minimum \(2^1\) to make the result a perfect Square
In that Case \(n^2 = 2^4 * 3^2\)
i.e. \(n = 2^2 * 3 = 12\)
Answer: Option
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