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2023 AMC 10B Problems/Problem 18

Revision as of 20:06, 15 November 2023 by Technodoggo (talk | contribs)

Problem

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.

Solution 1 (Guess and check + Contrapositive)

being revised ~Technodoggo

Solution 2

The equation given in the problem can be written as \[ 15 a + 14 b = c. \hspace{1cm} (1) \]

$\textbf{First, we prove that Statement I is not correct.}$

A counter example is $a = 1$ and $b = 3$. Thus, ${\rm gcd} (c, 210) = 3 \neq 1$.

$\textbf{Second, we prove that Statement III is correct.}$

First, we prove the ``if part.

Suppose ${\rm gcd}(a , 14) = 1$ and ${\rm gcd}(b, 15) = 1$. However, ${\rm gcd} (c, 210) \neq 1$.

Thus, $c$ must be divisible by at least one factor of 210. W.L.O.G, we assume $c$ is divisible by 2.

Modulo 2 on Equation (1), we get that $2 | a$. This is a contradiction with the condition that ${\rm gcd}(a , 14) = 1$. Therefore, the ``if part in Statement III is correct.

Second, we prove the ``only if part.

Suppose ${\rm gcd} (c, 210) \neq 1$. Because $210 = 14 \cdot 15$, there must be one factor of 14 or 15 that divides $c$. W.L.O.G, we assume there is a factor $q > 1$ of 14 that divides $c$. Because ${\rm gcd} (14, 15) = 1$, we have ${\rm gcd} (q, 15) = 1$. Modulo $q$ on Equation (1), we have $q | a$.

Because $q | 14$, we have ${\rm gcd}(a , 14) \geq q > 1$.

Analogously, we can prove that ${\rm gcd}(b , 15) > 1$.

$\textbf{Third, we prove that Statement II is correct.}$

This is simply a special case of the ``only if part of Statement III. So we omit the proof.

All analysis above imply $\boxed{\textbf{(E) II and III only}}.$

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

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