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21 May 2024, 01:28
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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0% (00:00) correct
100% (01:32) wrong based on 2 sessions
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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?
Indicate all such integers.
A. 12
B. 24
C. 36
D. 72
Indicate all such integers.
A. 12
B. 24
C. 36
D. 72
06 Jun 2024, 00:58
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Now here we are looking for a Set S of numbers n such that \(n^2\) is multiple of 24 and 108.
Lets take one such case Least Common Multiple of 108 and 24 is 216.
\(n^2\) = 216
n = 6\(\sqrt{6}\).
This is not an integer however if we multiply 216 by any number it is still divisible by 24 and 108. So we multiply by 6 and repat the above steps to find n. So n =36.
Now n= 36 is one of the number in set S.
Hence all the options that divide n =36 are correct. Hence option A and C.
Lets take one such case Least Common Multiple of 108 and 24 is 216.
\(n^2\) = 216
n = 6\(\sqrt{6}\).
This is not an integer however if we multiply 216 by any number it is still divisible by 24 and 108. So we multiply by 6 and repat the above steps to find n. So n =36.
Now n= 36 is one of the number in set S.
Hence all the options that divide n =36 are correct. Hence option A and C.
26 Jun 2024, 12:40
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Official Explanation
To determine which of the integers in the answer choices is a divisor of every positive integer n in S, you must first understand the integers that are in S. Note that in this question you are given information about \(n^2\), not about n itself. Therefore, you must use the information about to derive information about n.
The fact that \(n^2\) is a multiple of both 24 and 108 implies that \(n^2\) is a multiple of the least common multiple of 24 and 108. To determine the least common multiple of 24 and 108, factor 24 and 108 into prime factors as \((2^3)(3)\) and \((2^3)(3^3)\), respectively. Because these are prime factorizations, you can conclude that the least common multiple of 24 and 108 is \((2^3)(3^3)\).
Knowing that n^2 must be a multiple of \((2^3)(3^3)\) does not mean that every multiple of \((2^3)(3^3)\) is a possible value of \(n^2\), because \(n^2\) must be the square of an integer. The prime factorization of a square number must contain only even exponents. Thus, the least multiple of \((2^3)(3^3)\) that is a square is \((2^4)(3^4)\). This is the least possible value of \(n^2\), and so the least possible value of n is \((2^3)(3^3)\), or 36. Furthermore, since every value of \(n^2\) is a multiple of \((2^4)(3^4)\), the values of n are the positive multiples of 36; that is, S = {36, 72, 108, 144, 180, …}.
The question asks for integers that are divisors of every integer n in S, that is, divisors of every positive multiple of 36. Since Choice A, 12, is a divisor of 36, it is also a divisor of every multiple of 36. The same is true for Choice C, 36. Choices B and D, 24 and 72, are not divisors of 36, so they are not divisors of every integer in S.
The correct answer consists of Choices A and C.
To determine which of the integers in the answer choices is a divisor of every positive integer n in S, you must first understand the integers that are in S. Note that in this question you are given information about \(n^2\), not about n itself. Therefore, you must use the information about to derive information about n.
The fact that \(n^2\) is a multiple of both 24 and 108 implies that \(n^2\) is a multiple of the least common multiple of 24 and 108. To determine the least common multiple of 24 and 108, factor 24 and 108 into prime factors as \((2^3)(3)\) and \((2^3)(3^3)\), respectively. Because these are prime factorizations, you can conclude that the least common multiple of 24 and 108 is \((2^3)(3^3)\).
Knowing that n^2 must be a multiple of \((2^3)(3^3)\) does not mean that every multiple of \((2^3)(3^3)\) is a possible value of \(n^2\), because \(n^2\) must be the square of an integer. The prime factorization of a square number must contain only even exponents. Thus, the least multiple of \((2^3)(3^3)\) that is a square is \((2^4)(3^4)\). This is the least possible value of \(n^2\), and so the least possible value of n is \((2^3)(3^3)\), or 36. Furthermore, since every value of \(n^2\) is a multiple of \((2^4)(3^4)\), the values of n are the positive multiples of 36; that is, S = {36, 72, 108, 144, 180, …}.
The question asks for integers that are divisors of every integer n in S, that is, divisors of every positive multiple of 36. Since Choice A, 12, is a divisor of 36, it is also a divisor of every multiple of 36. The same is true for Choice C, 36. Choices B and D, 24 and 72, are not divisors of 36, so they are not divisors of every integer in S.
The correct answer consists of Choices A and C.
