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If n is a positive integer and n^2 is divisible by 72, then
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17 Aug 2020, 10:30
17 Aug 2020, 10:30
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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
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: If n is a positive integer and n^2 is divisible by 72, then
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17 Aug 2020, 10:41
17 Aug 2020, 10:41
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Carcass 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
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
Answer: B
Cheers,
Brent
General Discussion
Re: If n is a positive integer and n^2 is divisible by 72, then
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17 Aug 2020, 10:31
17 Aug 2020, 10:31
1
Expert Reply
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Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
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