Difference between revisions of "2023 AMC 10B Problems/Problem 18"

Revision as of 16:51, 15 November 2023

Suppose 𝑎, 𝑏, and 𝑐 are positive integers such that $\dfrac{a}{14}+\dfrac{b}{15}=\dfrac{c}{210}$ .

Which of the following statements are necessarily true?

I. If gcd(𝑎, 14) = 1 or gcd(𝑏, 15) = 1 or both, then gcd(𝑐, 210) = 1.

II. If gcd(𝑐, 210) = 1, then gcd(𝑎, 14) = 1 or gcd(𝑏, 15) = 1 or both.

III. gcd(𝑐, 210) = 1 if and only if gcd(𝑎, 14) = gcd(𝑏, 15) = 1.

$I.$ Try $a=3,b=5 => c = 17\cdot15$ which makes $\textbf{I}$ false. At this point, we can rule out answer A,B,C.

$II.$ A => B or C. equiv. ~B AND ~C => ~A. Let a = 14, b=15 (statisfying ~B and ~C). => C = 2*210. which is ~A.

$II$ is true.

So the answer is E. $\boxed{\textbf{(E) } II \text{ and } III \text{only}.}$ ~Technodoggo

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 Which of the following statements are necessarily true?
-I. If gcd(𝑎, 14) = 1 or gcd(𝑏, 15) = 1 or both, then gcd(𝑐, 21) = 1.
+I. If gcd(𝑎, 14) = 1 or gcd(𝑏, 15) = 1 or both, then gcd(𝑐, 210) = 1.
-II. If gcd(𝑐, 21) = 1, then gcd(𝑎, 14) = 1 or gcd(𝑏, 15) = 1 or both.
+II. If gcd(𝑐, 210) = 1, then gcd(𝑎, 14) = 1 or gcd(𝑏, 15) = 1 or both.
-III. gcd(𝑐, 21) = 1 if and only if gcd(𝑎, 14) = gcd(𝑏, 15) = 1.
+III. gcd(𝑐, 210) = 1 if and only if gcd(𝑎, 14) = gcd(𝑏, 15) = 1.
 == Solution (Guess and check + Contrapositive)==