AMC10 2023 B

AMC10 2023 B · Q18

AMC10 2023 B · Q18. It mainly tests Manipulating equations, GCD & LCM.

Suppose $a$, $b$, and $c$ are positive integers such that\[\frac{a}{14}+\frac{b}{15}=\frac{c}{210}.\]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 $\gcd(a,14)=\gcd(b,15)=1$.

假设$a$、$b$和$c$是正整数，使得\[\frac{a}{14}+\frac{b}{15}=\frac{c}{210}.\]以下哪些陈述必然成立？ I. 如果$\gcd(a,14)=1$或$\gcd(b,15)=1$或两者皆然，则$\gcd(c,210)=1$。 II. 如果$\gcd(c,210)=1$，则$\gcd(a,14)=1$或$\gcd(b,15)=1$或两者皆然。 III. $\gcd(c,210)=1$当且仅当$\gcd(a,14)=\gcd(b,15)=1$。

(A) \text{I, II, and III} \text{I, II, and III}

(B) \text{I only} \text{I only}

(D) \text{III only} \text{III only}

(E) \text{II and III only} \text{II and III only}

Answer

Correct choice: (E)

正确答案：（E）

Solution

We examine each of the conditions. The first condition is false. A simple counterexample is $a=3$ and $b=5$. The corresponding value of $c$ is $115$. Since $\gcd(3,14)=1$, condition $I$ would imply that $\gcd(c,210)=1.$ However, $\gcd(115,210)$ is clearly not $1$ (they share a common factor of $5$). Condition $I$ is false so that we can rule out choices $A,B,$ and $C$. We now decide between the two answer choices $D$ and $E$. What differs between them is the validity of condition $II$, so it suffices to check $II$ simply. We look at statement $II$'s contrapositive to prove it. The contrapositive states that if $\gcd(a,14)\neq1$ and $\gcd(b,15)\neq1$, then $\gcd(c,210)\neq1.$ In other words, if $a$ shares some common factor that is not $1$ with $14$ and $b$ shares some common factor that is not $1$ with $15$, then $c$ also shares a common factor that is not $1$ with $210$. Let's say that $a=a'\cdot n$, where $a'$ is a factor of $14$ not equal to $1$. (So $a'$ is the common factor.) We can rewrite the given equation as $15a+14b=c\implies15(a'n)+14b=c.$ We can express $14$ as $a'\cdot n'$, for some positive integer $n'$ (this $n'$ can be $1$). We can factor $a'$ out to get $a'(15n+bn')=c.$ Since all values in this equation are integers, $c$ must be divisible by $a'$. Since $a'$ is a factor of $14$, $a'$ must also be a factor of $210$, a multiple of $14$. Therefore, we know that $c$ shares a common factor with $210$ (which is $a'$), so $\gcd(c,210)\neq1$. This is what $II$ states, so therefore $II$ is true. Thus, our answer is $\boxed{\textbf{(E) }\text{II and III only}}.$

我们逐一考察每个条件。第一个条件是假的。一个简单的反例是$a=3$和$b=5$。对应的$c=115$。因为$\gcd(3,14)=1$，条件I会暗示$\gcd(c,210)=1$。但是，$\gcd(115,210)$明显不为$1$（它们有公因子$5$）。条件I是假的，因此排除选项$A,B$和$C$。现在在$D$和$E$之间选择。它们的不同在于条件II的有效性，因此只需检查II。我们考察语句II的对偶来证明它。对偶陈述：如果$\gcd(a,14)\neq1$且$\gcd(b,15)\neq1$，则$\gcd(c,210)\neq1$。换句话说，如果$a$与$14$有非$1$公因子且$b$与$15$有非$1$公因子，则$c$也与$210$有非$1$公因子。设$a=a'\cdot n$，其中$a'$是$14$的非$1$因子（即公因子）。我们可以改写给定方程为$15a+14b=c\implies15(a'n)+14b=c$。我们可以将$14$表示为$a'\cdot n'$，其中$n'$是正整数（$n'$可以是$1$）。我们可以提取$a'$得到$a'(15n+bn')=c$。因为方程中所有值为整数，$c$必须被$a'$整除。因为$a'$是$14$的因子，$a'$也必须是$210$的因子（$210$是$14$的倍数）。因此，$c$与$210$有公因子（即$a'$），所以$\gcd(c,210)\neq1$。这就是II陈述的内容，因此II是真。因此，答案为$\boxed{\textbf{(E) }\text{II and III only}}$。

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