Answered

Suppose that $\operatorname{GCD}(a, b) = \operatorname{GCD}(b, c) = 2$. Does that always imply that $\operatorname{GCD}(a, b, c) = 2$? Justify your answer.



Select the correct choice below.

AYes, the implication is true. The hypothesis implies that 2 is a common multiple of $a, b$, and $c$. If a number greater than 2 is a multiple of $a, b$, and $c$, then in particular that number would have to be a multiple of $a$ and $b$, which is impossible because $\operatorname{GCD}(a, b) = 2$.
BYes, the implication is true. The hypothesis implies that 2 is a common divisor of $a, b$, and $c$. If a number greater than 2 divided $a, b$, and $c$, then in particular that number would have to divide $a$ and $b$, which is impossible because $\operatorname{GCD}(a, b) = 2$.
CNo, the implication is not true. The hypothesis implies that 2 is a common multiple of $a, b$, and $c$. If a number greater than 2 is a multiple of $a, b$, and $c$, then in particular that number would have to be a multiple of $a$ and $b$, which is possible because $\operatorname{GCD}(a, b) = 2$.
DNo, the implication is not true. The hypothesis implies that 2 is a common divisor of $a, b$, and $c$. If a number greater than 2 divided $a, b$, and $c$, then in particular that number would have to divide $a$ and $b$, which is possible because $\operatorname{GCD}(a, b) = 2$.

Verified Answer :

  • The condition \(\gcd(a, b) = 2\) and \(\gcd(b, c) = 2\) implies that \(2\) is a common divisor of \(a, b,\) and \(c\).
  • Any common divisor \(d\) of \(a, b,\) and \(c\) must also be a common divisor of \(a\) and \(b\).
  • Since \(\gcd(a, b) = 2\), the largest possible common divisor of \(a\) and \(b\) is \(2\), so \(d \leq 2\).
  • Thus, the greatest common divisor \(\gcd(a, b, c)\) is exactly \(2\).
  • The correct choice is B.

\(\boxed{B}\)

Explanation

  1. Analysis of the Problem

The problem asks to determine if the condition \(\gcd(a, b) = 2\) and \(\gcd(b, c) = 2\) implies that \(\gcd(a, b, c) = 2\). Let's analyze the meaning of \(\gcd(a, b, c)\). It represents the greatest integer that divides \(a\), \(b\), and \(c\) simultaneously. If \(d = \gcd(a, b, c)\), then \(d\) must divide \(a, b,\) and \(c\).

  1. Logical Deduction of the GCD

Since \(2\) is given as a common divisor of \(a\) and \(b\), and \(2\) is also a common divisor of \(b\) and \(c\), it follows that \(2\) must be a common divisor of all three numbers \(a, b,\) and \(c\). Therefore, \(2 \leq \gcd(a, b, c)\). We must now check if a common divisor \(d\) larger than \(2\) can exist. If a common divisor \(d\) exists such that \(d > 2\), then \(d\) must divide \(a, b,\) and \(c\). In particular, \(d\) must divide both \(a\) and \(b\). By definition, if \(d\) divides both \(a\) and \(b\), it cannot be greater than the greatest common divisor of \(a\) and \(b\). We are given that \(\gcd(a, b) = 2\). Therefore, no integer \(d > 2\) can divide both \(a\) and \(b\). Consequently, no integer \(d > 2\) can divide all three of \(a, b,\) and \(c\).

  1. Conclusion and Selection of Option

Since \(2\) is a common divisor and no common divisor exists that is strictly greater than \(2\), the greatest common divisor \(\gcd(a, b, c)\) must be \(2\). This confirms that the implication is true. Comparing this logic with the given options, option B correctly states that 2 is a common divisor of \(a, b,\) and \(c\), and that if a larger number divided all three, it would contradict \(\gcd(a, b) = 2\).

Examples

Imagine you are organizing three different manufacturing schedules (A, B, and C) that repeat their maintenance cycles. Schedule A and B have maintenance every 2 weeks (GCD=2), and Schedule B and C have maintenance every 2 weeks (GCD=2). This structure ensures that all three schedules align their maintenance breaks every 2 weeks, and they don't have a more frequent shared maintenance interval (like every 4 weeks) because the shared interval between A and B is limited to 2 weeks. Therefore, the GCD of all three is 2, ensuring consistent coordination.

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