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MBA.Aspirant: Master | Next Rank: 500 Posts; Posts: 401; Joined: Tue May 24, 2011 1:14 amFollowed by: 5 members; Thanked: 37 times

by MBA.Aspirant » Mon Nov 21, 2011 11:35 am

Let S be the set of all positive integers n such that n^2 is a multiple of both 24 and 108. Which of the following integers are divisors of every integer n in S?

Check all that apply

A) 12
B) 24
C) 36
D) 72

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Source: Beat The GMAT — Problem Solving | Mon Nov 21, 2011 11:35 am

shankar.ashwin: Legendary Member; Posts: 966; Joined: Sat Jan 02, 2010 8:06 amFollowed by: 21 members; Thanked: 230 times

by shankar.ashwin » Mon Nov 21, 2011 12:08 pm

Consider the smallest number in the set, that is a multiple of 24 and 108 which is also a perfect square.

First common multiple of 24 and 108 (LCM) = 216, but 216 is not a perfect square.

216 = 2^3 * 3^3 (To make it a perfect square (even powers) we can multiply it by 2*3 )

All multiples of 216 will remain multiples of 24 and 108.

216*6 = 1296. (this is n^2)

Therefore n = 36. Only A (12) and C(36) are multiples

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Mon Nov 21, 2011 9:39 pm

MBA.Aspirant wrote:Let S be the set of all positive integers n such that n^2 is a multiple of both 24 and 108. Which of the following integers are divisors of every integer n in S?

Check all that apply

A) 12
B) 24
C) 36
D) 72

24 = 4 * 6 = 2Â² * 6, here we need a 6 to make it a perfect square, which will be 144, and âˆš144 = 12. So, 12, 14, 36, 48, 60, 72... satisfy the first condition.

108 = 3 * 4 * 9 = 3 * 4Â² * 3Â², here we need a 3 to make it a perfect square, which will be 324, and âˆš324 = 18. So, 18, 36, 54, 72,... satisfy the second condition.

Therefore, the numbers that satisfy both conditions are S = {36, 72, 108,...}
Hence, the integers that are divisors of every integer n in S are 12 and 36, which implies A and C.

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pemdas: Legendary Member; Posts: 1084; Joined: Fri Apr 15, 2011 2:33 pmFollowed by: 21 members; Thanked: 158 times

by pemdas » Mon Nov 21, 2011 11:08 pm

24 prime factorized as (2^3)*3
108 prime factorized as (2^2)*(3^3)

all positive integers (discard 0) n such that n^2 is multiple of 24 and 108

lcm (24,108) turned into the perfect square and factored by any whole number squared will suit here

lcm(24,108)=(2^3)(3^3) which can be turned into the perfect square, n^2 provided (2*3)^3 *(2*3) OR (2*3)^4=36^2=n^2
n=36. 36 factored by any number squared (except for 0, n must be positive the resultant) should be divisible by selected (correct) answer choices

A) 36 is divisible by 12
B) 36 is not divisible by 24 BUT 36 can be multiplied by 2 and be divisible by 24. However, not every integer n^2 will be multiple of 24 and 108, e.g. 36^2 is a multiple of 24 and 108 but 36 isn't divisible by 24 (condition every integer is not observed)
C) 36 is divisible by 36
D) 36 isn't divisible by 72

MBA.Aspirant wrote:Let S be the set of all positive integers n such that n^2 is a multiple of both 24 and 108. Which of the following integers are divisors of every integer n in S?

Check all that apply

A) 12
B) 24
C) 36
D) 72

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