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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MBA.Aspirant
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shankar.ashwin
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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
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
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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 = 2² * 6, here we need a 6 to make it a perfect square, which will be 144, and √144 = 12. So, 12, 14, 36, 48, 60, 72... satisfy the first condition.
108 = 3 * 4 * 9 = 3 * 4² * 3², here we need a 3 to make it a perfect square, which will be 324, and √324 = 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
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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
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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