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Largest positive divisor of positive integer n

Expert replies
by gander123 » Fri Jan 04, 2013 4:30 am
Hoping that noone here has ever considered me an "Instigator" (Knewton), I would like to discuss the following problem with you.

GMAT Official Guide Quantitative Review (2nd ed.):

Q 169 Problem Sovling

"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

Correct Answer: B

Official Answer Explanation:

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 = 2*m^2 for some positive integer m in order for 72k to be a perfect square. Then, n^2 = 72k = (2^3)*(3^2)*(2*m^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 largest positive integer that must divide n is 12.

Please discuss:

The boldfaced sentence concerns my question. Why dont we just say that k = 2 so that 72k = 144 which is indeed a perfect square of 12^2 ?? Taking k = 2 instead of 2m^2 would yield the same result namely n^2= [(2^2)*(3)]^2 and thus n = (2^2)*(3) from which we would still be able to conclude that 12 is the largest possible integer that must divide n.

To me, it seemed that the introduction of m^2 did not contribute anything in the finding of the solution. Why did they used it though?

Kind regards,

Tobi
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Source: — Problem Solving |

by Param800 » Fri Jan 04, 2013 4:51 am
Hey Tobi,

I did this question like this, maybe it will help you to understand your concern.

So...we are given that N^2 is divisible by 72 and N is an integer.

And we know that 72=2*6*6

Since N^2 is completely divisible by 72, it should yield an integer as its result on division.
It can only yield an integer if we multiply 72 by 2...that is if N^2 = 2* 72 = 2*2 *6*6

which yields N as 2*6 = 12
We can also multiply it by any other pair of integers like 4*4 , 5*5...etc but basically N will have 12 as a common factor in all these numbers.

So, since the question is asking us what is the largest positive integer that must divide N ... we know that it should be 12.

Hence answer B.

gander123 wrote:Hoping that noone here has ever considered me an "Instigator" (Knewton), I would like to discuss the following problem with you.

GMAT Official Guide Quantitative Review (2nd ed.):

Q 169 Problem Sovling

"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

Correct Answer: B

Official Answer Explanation:

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 = 2*m^2 for some positive integer m in order for 72k to be a perfect square. Then, n^2 = 72k = (2^3)*(3^2)*(2*m^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 largest positive integer that must divide n is 12.

Please discuss:

The boldfaced sentence concerns my question. Why dont we just say that k = 2 so that 72k = 144 which is indeed a perfect square of 12^2 ?? Taking k = 2 instead of 2m^2 would yield the same result namely n^2= [(2^2)*(3)]^2 and thus n = (2^2)*(3) from which we would still be able to conclude that 12 is the largest possible integer that must divide n.

To me, it seemed that the introduction of m^2 did not contribute anything in the finding of the solution. Why did they used it though?

Kind regards,

Tobi
Join the discussion

by gander123 » Fri Jan 04, 2013 6:57 am
Hey Param
It can only yield an integer if we multiply 72 by 2...that is if N^2 = 2* 72 = 2*2 *6*6
What do you mean by "It" at the beginning...

Perhaps I'm driving the question to far....

Cheers,

Tobi
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by Param800 » Fri Jan 04, 2013 8:50 am
Hey Tobi,

I meant to say N^2... since questions says " n^2 is divisible by 72 "...which means the quotient should be an integer..consider this example 4 is divisible by 2...which means quotient is 2..ie an integer...we can't say that the quotient will be a non-integer number because then it will not be completely divisible.

Actually I think it would be a good idea, if the question mentioned this... n^2 is evenly divisible by 72 ...which clearly mentions that the quotient should be an integer.
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by GMATGuruNY » Fri Jan 04, 2013 8:58 am
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
To determine the largest positive integer that MUST be a factor of n, we first must determine the LEAST possible value of n.

Since 72 = 2³ * 3², n² must be divisible by 2³ and 3².
Thus, the LEAST possible value of n = 2*2*3, so that n² = (2*2*3)² = 2� * 3², which is divisible by 2³ and 3².
(If n = 2*3, then n² = (2*3)² = 2² * 3², which is NOT divisible by 2³.)

If n = 2*2*3 = 12 -- the LEAST value that n could be -- then the greatest positive integer that MUST divide into n is 12.

The correct answer is B.
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by buzzdeepak » Fri Jan 04, 2013 3:01 pm
Not to get too technical, but "evenly divisible" and "divisible" mean the same i.e., no remainders / quotient should be integers.

Param800 wrote:Hey Tobi,

I meant to say N^2... since questions says " n^2 is divisible by 72 "...which means the quotient should be an integer..consider this example 4 is divisible by 2...which means quotient is 2..ie an integer...we can't say that the quotient will be a non-integer number because then it will not be completely divisible.

Actually I think it would be a good idea, if the question mentioned this... n^2 is evenly divisible by 72 ...which clearly mentions that the quotient should be an integer.
Join the discussion