| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve the problem, we need to analyze the given equation and the statements provided. The equation is: \[ \frac{a}{14} + \frac{b}{15} = \frac{c}{210} \] First, let's rewrite the equation in a more manageable form. We can find a common denominator for the fractions on the left-hand side: \[ \frac{15a + 14b}{210} = \frac{c}{210} \] Since the denominators are the same, we can equate the numerators: \[ 15a + 14b = c \] Now, let's analyze each statement: ### Statement I "If \(\gcd(a, 14) = 1\) or \(\gcd(b, 15) = 1\) or both, then \(\gcd(c, 210) = 1\)." To check this, consider the prime factorizations: - \(14 = 2 \times 7\) - \(15 = 3 \times 5\) - \(210 = 2 \times 3 \times 5 \times 7\) If \(\gcd(a, 14) = 1\), then \(a\) is not divisible by 2 or 7. If \(\gcd(b, 15) = 1\), then \(b\) is not divisible by 3 or 5. However, this does not necessarily mean that \(c = 15a + 14b\) will be coprime with 210. For example, if \(a = 1\) and \(b = 5\), then: \[ c = 15 \times 1 + 14 \times 5 = 15 + 70 = 85 \] Here, \(\gcd(85, 210) = 5\), not 1. Thus, Statement I is false. ### Statement II "If \(\gcd(c, 210) = 1\), then \(\gcd(a, 14) = 1\) or \(\gcd(b, 15) = 1\) or both." If \(\gcd(c, 210) = 1\), then \(c\) is not divisible by 2, 3, 5, or 7. Since \(c = 15a + 14b\), for \(c\) to be coprime with 210, neither \(a\) nor \(b\) can introduce any of these factors into \(c\). Therefore, at least one of \(\gcd(a, 14) = 1\) or \(\gcd(b, 15) = 1\) must hold. Thus, Statement II is true. ### Statement III "\(\gcd(c, 210) = 1\) if and only if \(\gcd(a, 14) = \gcd(b, 15) = 1\)." This statement is a stronger version of Statement II. If \(\gcd(c, 210) = 1\), then both \(\gcd(a, 14) = 1\) and \(\gcd(b, 15) = 1\) must hold. Conversely, if both \(\gcd(a, 14) = 1\) and \(\gcd(b, 15) = 1\), then \(c\) will be coprime with 210. Thus, Statement III is true. Based on the analysis, the correct statements are II and III. Therefore, the answer is: \[ \boxed{E} \] |
E |
| deepseek-r1 | Given the equation \(\frac{a}{14} + \frac{b}{15} = \frac{c}{210}\) where \(a\), \(b\), and \(c\) are positive integers, we start by rewriting the equation as: \[ 15a + 14b = c \] We need to determine the validity of the statements I, II, and III regarding the greatest common divisors (gcd). **Statement I**: If \(\gcd(a, 14) = 1\) or \(\gcd(b, 15) = 1\) or both, then \(\gcd(c, 210) = 1\). - **Analysis**: We found counterexamples where \(\gcd(a, 14) = 1\) or \(\gcd(b, 15) = 1\) does not necessarily result in \(\gcd(c, 210) = 1\). For instance, if \(a = 1\) (coprime to 14) and \(b = 15\) (not coprime to 15), then \(c = 15 \cdot 1 + 14 \cdot 15 = 225\), and \(\gcd(225, 210) = 15\). Thus, Statement I is false. **Statement II**: If \(\gcd(c, 210) = 1\), then \(\gcd(a, 14) = 1\) or \(\gcd(b, 15) = 1\) or both. - **Analysis**: If \(\gcd(c, 210) = 1\), \(c\) must not be divisible by 2, 3, 5, or 7. We showed that this implies \(a\) must be coprime to 14 and \(b\) must be coprime to 15. Hence, both \(\gcd(a, 14) = 1\) and \(\gcd(b, 15) = 1\) must hold, making the "or" condition true. Thus, Statement II is true. **Statement III**: \(\gcd(c, 210) = 1\) if and only if \(\gcd(a, 14) = \gcd(b, 15) = 1\). - **Analysis**: We confirmed that if both \(\gcd(a, 14) = 1\) and \(\gcd(b, 15) = 1\), then \(c\) is coprime to 210. Conversely, if \(\gcd(c, 210) = 1\), both conditions must hold. Thus, Statement III is true. **Conclusion**: Statements II and III are necessarily true, but Statement I is not. Therefore, the correct answer is: \[ \boxed{EEEEE} \] |
E |
| gpto1 | The answer is **(E) II and III only**. Here's the reasoning: From the equation \(\frac{a}{14} + \frac{b}{15} = \frac{c}{210}\), multiplying both sides by 210 simplifies to \(15a + 14b = c\). Statement I: "If \(\gcd(a,14) = 1\) or \(\gcd(b,15) = 1\) or both, then \(\gcd(c,210) = 1\)." - This statement is **false** because, even if \(\gcd(a,14) = 1\) or \(\gcd(b,15) = 1\), \(c\) may still share common factors with 210 (like 3 or 5), so we can't guarantee \(\gcd(c,210) = 1\). Statement II: "If \(\gcd(c,210) = 1\), then \(\gcd(a,14) = 1\) or \(\gcd(b,15) = 1\) or both." - This statement is **true**. If \(c\) shares no common factors with 210, then both \(\gcd(a,14) = 1\) and \(\gcd(b,15) = 1\), so at least one of them must be 1. Statement III: "\(\gcd(c,210) = 1\) if and only if \(\gcd(a,14) = \gcd(b,15) = 1\)." - This statement is **true**. We have shown the "if" and "only if" directions: If both \(\gcd(a,14) = 1\) and \(\gcd(b,15) = 1\), then \(\gcd(c,210) = 1\), and vice versa. Therefore, the correct choice is **(E) II and III only**. **Answer:** EEEEE |
E |
| gpt-4o-2024-08-06 | To solve this problem, we need to analyze the given equation: \[ \frac{a}{14} + \frac{b}{15} = \frac{c}{210}. \] First, we find a common denominator for the left-hand side: \[ \frac{a}{14} + \frac{b}{15} = \frac{15a + 14b}{210}. \] Thus, we have: \[ \frac{15a + 14b}{210} = \frac{c}{210}. \] This implies: \[ 15a + 14b = c. \] Now, let's analyze each statement: **Statement I:** If \(\gcd(a,14)=1\) or \(\gcd(b,15)=1\) or both, then \(\gcd(c,210)=1\). - If \(\gcd(a,14) = 1\), then \(a\) is not divisible by 2 or 7. - If \(\gcd(b,15) = 1\), then \(b\) is not divisible by 3 or 5. Since \(c = 15a + 14b\), we need to check if \(c\) can be divisible by any of the prime factors of 210, which are 2, 3, 5, and 7. - If \(\gcd(a,14) = 1\), then \(15a\) is not divisible by 2 or 7, and \(14b\) is not divisible by 2 or 7. Thus, \(c\) is not divisible by 2 or 7. - If \(\gcd(b,15) = 1\), then \(15a\) is not divisible by 3 or 5, and \(14b\) is not divisible by 3 or 5. Thus, \(c\) is not divisible by 3 or 5. Therefore, if either \(\gcd(a,14) = 1\) or \(\gcd(b,15) = 1\), then \(\gcd(c,210) = 1\). Statement I is true. **Statement II:** If \(\gcd(c,210)=1\), then \(\gcd(a,14)=1\) or \(\gcd(b,15)=1\) or both. - If \(\gcd(c,210) = 1\), then \(c\) is not divisible by 2, 3, 5, or 7. - If \(\gcd(a,14) \neq 1\), then \(a\) is divisible by 2 or 7. - If \(\gcd(b,15) \neq 1\), then \(b\) is divisible by 3 or 5. If both \(\gcd(a,14) \neq 1\) and \(\gcd(b,15) \neq 1\), then \(c = 15a + 14b\) would be divisible by at least one of 2, 3, 5, or 7, contradicting \(\gcd(c,210) = 1\). Therefore, at least one of \(\gcd(a,14) = 1\) or \(\gcd(b,15) = 1\) must hold. Statement II is true. **Statement III:** \(\gcd(c,210)=1\) if and only if \(\gcd(a,14)=\gcd(b,15)=1\). - We have shown that if \(\gcd(a,14) = 1\) or \(\gcd(b,15) = 1\), then \(\gcd(c,210) = 1\). - However, if \(\gcd(c,210) = 1\), it is not necessary for both \(\gcd(a,14) = 1\) and \(\gcd(b,15) = 1\) to hold; only one of them needs to be true. Therefore, Statement III is false. The correct answer is that Statements I and II are true, but Statement III is false. Thus, the answer is \(\textbf{(C)}\). CCCCCC |
E |
| DeepSeek-Coder-V2-Lite-Instruct | To solve the given problem, we need to analyze the equation \(\frac{a}{14} + \frac{b}{15} = \frac{c}{210}\) and evaluate the truth of the given statements. First, let's find a common denominator for the fractions on the left-hand side: \[ \frac{a}{14} + \frac{b}{15} = \frac{15a + 14b}{210} \] Thus, the equation becomes: \[ \frac{15a + 14b}{210} = \frac{c}{210} \] By equating the numerators, we get: \[ 15a + 14b = c \] Now, let's analyze each statement: **Statement I:** If \(\gcd(a, 14) = 1\) or \(\gcd(b, 15) = 1\) or both, then \(\gcd(c, 210) = 1\). To test this, consider a counterexample. Let \(a = 3\) and \(b = 5\): \[ 15 \cdot 3 + 14 \cdot 5 = 45 + 70 = 115 \] Here, \(\gcd(3, 14) = 1\) and \(\gcd(5, 15) = 1\), but \(\gcd(115, 210) = 5 \neq 1\). Therefore, Statement I is false. **Statement II:** If \(\gcd(c, 210) = 1\), then \(\gcd(a, 14) = 1\) or \(\gcd(b, 15) = 1\) or both. If \(\gcd(c, 210) = 1\), then \(c\) must be coprime with 210. Since \(c = 15a + 14b\), for \(\gcd(c, 210) = 1\) to hold, at least one of \(a\) or \(b\) must be such that \(15a + 14b\) is coprime with 210. This implies that at least one of \(a\) or \(b\) must be coprime with 14 or 15, respectively. Therefore, Statement II is true. **Statement III:** \(\gcd(c, 210) = 1\) if and only if \(\gcd(a, 14) = \(\gcd(b, 15) = 1\). This statement is a combination of the "if" part of Statement II and the "only if" part of Statement I. The "if" part is true (as shown in Statement II), but the "only if" part is false (as shown in Statement I). Therefore, Statement III is false. Based on the analysis, only Statement II is true. Thus, the correct answer is: \[ \boxed{C} \] The final answer is: \[ \boxed{C} \] |
E |
| Meta-Llama-3.1-405B-Instruct-Turbo | ## Step 1: Understand the given equation and the conditions The given equation is $\frac{a}{14}+\frac{b}{15}=\frac{c}{210}$, where $a$, $b$, and $c$ are positive integers. We need to analyze the statements about the greatest common divisors (gcd) of $a$, $b$, $c$ with $14$, $15$, and $210$. ## Step 2: Analyze statement I Statement I claims that if $\gcd(a,14)=1$ or $\gcd(b,15)=1$ or both, then $\gcd(c,210)=1$. We need to evaluate the validity of this statement. ## Step 3: Analyze statement II Statement II claims that if $\gcd(c,210)=1$, then $\gcd(a,14)=1$ or $\gcd(b,15)=1$ or both. We need to evaluate the validity of this statement. ## Step 4: Analyze statement III Statement III claims that $\gcd(c,210)=1$ if and only if $\gcd(a,14)=\gcd(b,15)=1$. We need to evaluate the validity of this statement. ## Step 5: Examine the equation for statement I We can rewrite the equation as $15a + 14b = c$. If $\gcd(a,14)=1$, then $a$ has no common factors with $14$. However, this does not necessarily mean that $c$ has no common factors with $210$. The same applies to $\gcd(b,15)=1$. Therefore, statement I is not necessarily true. ## Step 6: Examine the equation for statement II If $\gcd(c,210)=1$, then $c$ has no common factors with $210$. This means that $15a + 14b$ has no common factors with $210$. However, this does not necessarily mean that $a$ has no common factors with $14$ or $b$ has no common factors with $15$. Therefore, statement II is not necessarily true. ## Step 7: Examine the equation for statement III Statement III is a combination of statements I and II. Since we have found that statements I and II are not necessarily true, statement III is also not necessarily true. ## Step 8: Determine the correct answer Based on the analysis, none of the statements are necessarily true. However, this option is not available in the answer choices. The closest answer is $\textbf{(D)}~\text{III only}$, but since we cannot determine the correct multiple-choice answer with certainty, we will take our best guess. The final answer is: $\boxed{\text{DDDDD}}$ |
E |
| Meta-Llama-3.1-70B-Instruct-Turbo | ## Step 1: Understand the problem We are given an equation involving the variables $a$, $b$, and $c$, which are positive integers. The equation is $\frac{a}{14}+\frac{b}{15}=\frac{c}{210}$. We need to determine which of the given statements are necessarily true based on this equation. ## Step 2: Analyze statement I Statement I claims that if $\gcd(a,14)=1$ or $\gcd(b,15)=1$ or both, then $\gcd(c,210)=1$. To verify this, we can start by finding the least common multiple (LCM) of $14$ and $15$, which is $210$. Multiplying both sides of the equation by $210$ gives us $15a + 14b = c$. If $\gcd(a,14)=1$ or $\gcd(b,15)=1$ or both, then $c$ will not share any common factors with $210$, resulting in $\gcd(c,210)=1$. ## Step 3: Analyze statement II Statement II claims that if $\gcd(c,210)=1$, then $\gcd(a,14)=1$ or $\gcd(b,15)=1$ or both. We can use a counterexample to test this statement. Let $a=2$, $b=7$, and $c=1$. Substituting these values into the equation, we get $\frac{2}{14}+\frac{7}{15}=\frac{1}{210}$. Here, $\gcd(c,210)=1$, but $\gcd(a,14) \neq 1$ and $\gcd(b,15) \neq 1$. Therefore, statement II is not necessarily true. ## Step 4: Analyze statement III Statement III claims that $\gcd(c,210)=1$ if and only if $\gcd(a,14)=\gcd(b,15)=1$. We have already shown that if $\gcd(a,14)=1$ or $\gcd(b,15)=1$ or both, then $\gcd(c,210)=1$. However, the converse is not necessarily true, as demonstrated by the counterexample in step 3. ## Step 5: Determine the correct answer Based on our analysis, only statement I is necessarily true. The final answer is: $\boxed{B}$ |
E |
| Qwen2-72B-Instruct | To solve this problem, we need to analyze each statement individually based on the given equation \(\frac{a}{14} + \frac{b}{15} = \frac{c}{210}\). First, let's simplify the equation by finding a common denominator for the fractions on the left-hand side. The least common multiple (LCM) of 14 and 15 is 210, so we can rewrite the equation as: \[15a + 14b = c\] Now, let's analyze each statement: **Statement I**: If \(\gcd(a,14)=1\) or \(\gcd(b,15)=1\) or both, then \(\gcd(c,210)=1\). This statement is not necessarily true. Consider the case where \(a=1\) and \(b=1\). Here, \(\gcd(a,14)=1\) and \(\gcd(b,15)=1\), but \(c=29\), which does not imply that \(\gcd(c,210)=1\) since 29 is prime and does not divide evenly into 210, but it doesn't mean that any number satisfying the condition will always result in \(\gcd(c,210)=1\). This statement is false because there could be cases where \(c\) shares a common factor with 210 even when \(a\) and \(b\) do not share factors with their respective denominators. **Statement II**: If \(\gcd(c,210)=1\), then \(\gcd(a,14)=1\) or \(\gcd(b,15)=1\) or both. This statement is true. If \(\gcd(c,210)=1\), then \(c\) does not share any factors with 210. Since \(c=15a+14b\), if \(a\) and \(b\) shared any factors with 14 and 15 respectively, those factors would also be factors of \(c\), contradicting the assumption that \(\gcd(c,210)=1\). Therefore, at least one of \(a\) or \(b\) must not share any factors with their respective denominators, or both must not share factors. **Statement III**: \(\gcd(c,210)=1\) if and only if \(\gcd(a,14)=\gcd(b,15)=1\). This statement is false. As shown in the analysis of Statement I, it is possible for \(\gcd(c,210)\neq1\) even when \(\gcd(a,14)=1\) and \(\gcd(b,15)=1\). Conversely, it is also possible for \(\gcd(c,210)=1\) without \(\gcd(a,14)=\gcd(b,15)=1\), as long as the factors of \(a\) and \(b\) do not contribute to a common factor between \(c\) and 210. Therefore, the correct answer is **(E)**, which corresponds to **II** being true. However, since the question asks for the answer to be duplicated five times, the final answer is EEEEE. However, based on the options provided and the analysis, the correct answer should be interpreted as "II and III only" being evaluated, but since III is incorrect, the only necessarily true statement is II. Thus, the correct response to the question format would indeed be EEEEE, acknowledging that only Statement II is necessarily true. |
E |
| baichuan4 | To solve this problem, we need to analyze the given equation and its implications on the greatest common divisors (gcd) of the variables involved. The equation is: $$\frac{a}{14}+\frac{b}{15}=\frac{c}{210}$$ First, let's find a common denominator for the left side of the equation. The least common multiple (lcm) of 14 and 15 is 210. Therefore, we can rewrite the equation as: $$\frac{15a}{210}+\frac{14b}{210}=\frac{c}{210}$$ Next, we can combine the fractions on the left side since they have the same denominator: $$\frac{15a+14b}{210}=\frac{c}{210}$$ Since the denominators are the same, we can equate the numerators: $$15a+14b=c$$ Now, let's analyze each statement: I. If $\gcd(a,14)=1$ or $\gcd(b,15)=1$ or both, then $\gcd(c,210)=1$. If $\gcd(a,14)=1$, then $a$ is relatively prime to 14, which means $a$ does not share any prime factors with 14. Similarly, if $\gcd(b,15)=1$, then $b$ is relatively prime to 15. Since $c=15a+14b$, $c$ will be a sum of numbers that do not share prime factors with 14 and 15, respectively. Therefore, $c$ will not share any prime factors with 210, which is the product of 14 and 15. Thus, $\gcd(c,210)=1$. So, statement I is true. II. If $\gcd(c,210)=1$, then $\gcd(a,14)=1$ or $\gcd(b,15)=1$ or both. This statement is not necessarily true. Consider the case where $a$ is a multiple of 14 and $b$ is a multiple of 15, but their sum $c$ is relatively prime to 210. This can happen if the shared prime factors of $a$ and $b$ "cancel out" in the sum $c$. For example, if $a=14$ and $b=15$, then $c=15\cdot14+14\cdot15=210+210=420$, but $\gcd(420,210)=210\neq1$. So, statement II is false. III. $\gcd(c,210)=1$ if and only if $\gcd(a,14)=\gcd(b,15)=1$. This statement is not necessarily true either. The "if" part is true based on the argument for statement I, but the "only if" part is not true because it's possible for $c$ to be relatively prime to 210 even if one of $a$ or $b$ is not relatively prime to its respective number. For example, if $a=14$ and $b=1$, then $c=15\cdot14+14\cdot1=210+14=224$, and $\gcd(224,210)=1$, but $\gcd(14,14)\neq1$. So, statement III is false. Based on the analysis, only statement I is true. Therefore, the correct answer is: $\textbf{(B)}$ BBBBB |
E |
| doubao | C First, simplify the equation: $\frac{a}{14}+\frac{b}{15}=\frac{15a+14b}{210}=\frac{c}{210}$, so $15a+14b=c$. For statement I, if $\gcd(a,14)=1$ or $\gcd(b,15)=1$ or both, it doesn't necessarily mean $\gcd(c,210)=1$. For example, $a=1, b=15$ gives $c=30$ and $\gcd(c,210) \neq 1$. So statement I is false. For statement II, if $\gcd(c,210)=1$, then $c$ has no common factors with 210. Since $c=15a+14b$, if $a$ had a common factor with 14 or $b$ had a common factor with 15 that factor would also be in $c$. So $\gcd(a,14)=1$ or $\gcd(b,15)=1$ or both must be true. Statement II is true. For statement III, the "if and only if" is too strong. Just because $\gcd(c,210)=1$ doesn't mean both $\gcd(a,14)$ and $\gcd(b,15)$ are 1. And vice versa. So statement III is false. So the answer is C. |
E |
| gpto1 | The answer is **(E) II and III only**. Here's the reasoning: From the equation \(\frac{a}{14} + \frac{b}{15} = \frac{c}{210}\), multiplying both sides by 210 simplifies to \(15a + 14b = c\). Statement I: "If \(\gcd(a,14) = 1\) or \(\gcd(b,15) = 1\) or both, then \(\gcd(c,210) = 1\)." - This statement is **false** because, even if \(\gcd(a,14) = 1\) or \(\gcd(b,15) = 1\), \(c\) may still share common factors with 210 (like 3 or 5), so we can't guarantee \(\gcd(c,210) = 1\). Statement II: "If \(\gcd(c,210) = 1\), then \(\gcd(a,14) = 1\) or \(\gcd(b,15) = 1\) or both." - This statement is **true**. If \(c\) shares no common factors with 210, then both \(\gcd(a,14) = 1\) and \(\gcd(b,15) = 1\), so at least one of them must be 1. Statement III: "\(\gcd(c,210) = 1\) if and only if \(\gcd(a,14) = \gcd(b,15) = 1\)." - This statement is **true**. We have shown the "if" and "only if" directions: If both \(\gcd(a,14) = 1\) and \(\gcd(b,15) = 1\), then \(\gcd(c,210) = 1\), and vice versa. Therefore, the correct choice is **(E) II and III only**. **Answer:** EEEEE |
E |
| kimi | To solve this problem, we start by manipulating the given equation: \[ \frac{a}{14} + \frac{b}{15} = \frac{c}{210} \] First, we find a common denominator for the fractions on the left-hand side. The least common multiple of 14 and 15 is 210. Thus, we rewrite the equation as: \[ \frac{15a + 14b}{210} = \frac{c}{210} \] Since the denominators are equal, we can equate the numerators: \[ 15a + 14b = c \] We need to analyze the statements given in the problem in terms of the greatest common divisors (gcd) of \(a\), \(b\), and \(c\) with their respective coefficients. ### Statement I: If \(\gcd(a,14)=1\) or \(\gcd(b,15)=1\) or both, then \(\gcd(c,210)=1\). To determine this, we consider the prime factorization of 210: \[ 210 = 2 \times 3 \times 5 \times 7 \] We need to check if \(c\) can share any of these prime factors with 210 under the given conditions. - If \(\gcd(a,14) = 1\), then \(a\) is not divisible by 2 or 7. - If \(\gcd(b,15) = 1\), then \(b\) is not divisible by 3 or 5. Since \(c = 15a + 14b\), we analyze the possible prime factors of \(c\): - If \(a\) is not divisible by 2 or 7, and \(b\) is not divisible by 3 or 5, then \(15a\) is not divisible by 2 or 7, and \(14b\) is not divisible by 3 or 5. - Therefore, \(c = 15a + 14b\) cannot be divisible by 2, 3, 5, or 7. Thus, \(\gcd(c,210) = 1\). Statement I is true. ### Statement II: If \(\gcd(c,210)=1\), then \(\gcd(a,14)=1\) or \(\gcd(b,15)=1\) or both. Assume \(\gcd(c,210) = 1\). We need to show that either \(\gcd(a,14) = 1\) or \(\gcd(b,15) = 1\) or both. - If \(\gcd(a,14) \neq 1\), then \(a\) is divisible by either 2 or 7. - If \(\gcd(b,15) \neq 1\), then \(b\) is divisible by either 3 or 5. If both \(\gcd(a,14) \neq 1\) and \(\gcd(b,15) \neq 1\), then \(a\) is divisible by 2 or 7, and \(b\) is divisible by 3 or 5. This would make \(c = 15a + 14b\) divisible by at least one of 2, 3, 5, or 7, contradicting \(\gcd(c,210) = 1\). Therefore, either \(\gcd(a,14) = 1\) or \(\gcd(b,15) = 1\) or both. Statement II is true. ### Statement III: \(\gcd(c,210)=1\) if and only if \(\gcd(a,14)=\gcd(b,15)=1\). We have already shown that if \(\gcd(a,14) = 1\) or \(\gcd(b,15) = 1\) or both, then \(\gcd(c,210) = 1\). However, we need to check if the converse is also true. - If \(\gcd(c,210) = 1\), then neither \(a\) nor \(b\) can share any prime factors with 14 or 15, respectively. - This implies \(\gcd(a,14) = 1\) and \(\gcd(b,15) = 1\). Thus, \(\gcd(c,210) = 1\) if and only if \(\gcd(a,14) = 1\) and \(\gcd(b,15) = 1\). Statement III is true. Since all three statements are true, the correct answer is: \[ \boxed{\ |
E |
| minimax | Let's analyze each statement: I. If $\gcd(a,14)=1$ or $\gcd(b,15)=1$ or both, then $\gcd(c,210)=1$. This statement is true. The prime factorization of 14 is $2 \cdot 7$, and the prime factorization of 15 is $3 \cdot 5$. The least common multiple (LCM) of 14 and 15 is their product divided by their greatest common divisor (GCD), which is 1 since 14 and 15 are coprime. Therefore, the LCM of 14 and 15 is $2 \cdot 3 \cdot 5 \cdot 7 = 210$. The fraction $\frac{a}{14}+\frac{b}{15}$ can be rewritten as $\frac{a \cdot 15}{210}+\frac{b \cdot 14}{210}=\frac{15a + 14b}{210}$. If either $a$ or $b$ is coprime with its denominator (14 or 15, respectively), then $ |
E |
| wenxinyiyan | We begin by expressing the given equation in a more manageable form. Multiplying both sides by $210 \cdot 14 \cdot 15 = 44100$, we get \[15 \cdot 210a + 14 \cdot 210b = 14 \cdot 15c,\]which simplifies to \[3150a + 2940b = 210c.\]Now, we consider each statement individually. I. If $\gcd(a,14)=1$ or $\gcd(b,15)=1$ or both, then $\gcd(c,210)=1$. To prove this statement, we assume the contrary: that $\gcd(c, 210) \neq 1$. This means $c$ has a common factor with $210$ other than $1$. Since $210 = 2 \cdot 3 \cdot 5 \cdot 7$, the possible prime factors of $\gcd(c, 210)$ are $2$, $3$, $5$, and $7$. Suppose $\gcd(a, 14) = 1$ and $\gcd(c, 210) \neq 1$. If $c$ is divisible by any of $2$, $3$, $5$, or $7$, then the left side of the equation ($3150a + 2940b$) will not be divisible by that same prime, contradicting the equality. Similarly, if $\gcd(b, 15) = 1$ and $\gcd(c, 210) \neq 1$, we reach a contradiction. Therefore, if $\gcd(a,14)=1$ or $\gcd(b,15)=1$ or both, then $\gcd(c,210)$ must be $1$. Statement I is true. II. If $\gcd(c,210)=1$, then $\gcd(a,14)=1$ or $\gcd(b,15)=1$ or both. We prove this by contradiction as well. Assume $\gcd(c, 210) = 1$ but both $\gcd(a, 14) \neq 1$ and $\gcd(b, 15) \neq 1$. If $\gcd(a, 14) \neq 1$, then $a$ must have a common factor with $14$ other than $1$, which means $a$ is divisible by $2$ or $7$. Similarly, if $\gcd(b, 15) \neq 1$, then $b$ is divisible by $3$ or $5$. In either case, the left side of the equation ($3150a + 2940b$) will be divisible by at least one of $2$, $3$, $5$, or $7$, while the right side ($210c$) will not, contradicting the equality. Therefore, if $\gcd(c,210)=1$, then at least one of $\gcd(a,14)$ or $\gcd(b,15)$ must be $1$. Statement II is true. III. $\gcd(c,210)=1$ if and only if $\gcd(a,14)=\gcd(b,15)=1$. The "if" part follows directly from Statements I and II. If $\gcd(a,14)=\gcd(b,15)=1$, then by Statement I, $\gcd(c,210)=1$. However, the "only if" part is not true. It is possible for $\gcd(c,210)=1$ even if only one of $\gcd(a,14)$ or $\gcd(b,15)$ is $1$, as demonstrated in the proofs of Statements I and II. Therefore, Statement III is false. Since Statements I and II are true and Statement III is false, the correct answer is $\textbf{(C)}~\text{I and II only}$. Hence, the final answer string is: CCCCC. |
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| zhipu | 1. **Start by clearing the denominators**: Multiply through the equation $\frac{a}{14} + \frac{b}{15} = \frac{c}{210}$ by $210$ (the least common multiple of $14$, $15$, and $210$) to get: \[ 15a + 14b = c. \] 2. **Analyze Statement I**: Suppose $\gcd(a, 14) = 1$ or $\gcd(b, 15) = 1$ or both. We need to check if this implies $\gcd(c, 210) = 1$. - If $\gcd(a, 14) = 1$, then $a$ is not divisible by $2$ or $7$. - If $\gcd(b, 15) = 1$, then $b$ is not divisible by $3$ or $5$. - Since $c = 15a + 14b$, we need to check if $c$ can be divisible by any of $2$, $3$, $5$, or $7$ under these conditions. - If $a$ is not divisible by $2$ or $7$, and $b$ is not divisible by $3$ or $5$, then $c$ cannot be divisible by $2$, $3$, $5$, or $7$ because: - $15a$ is not divisible by $2$ or $7$ (since $15$ is not divisible by $2$ or $7$ and $a$ is not divisible by $2$ or $7$), - $14b$ is not divisible by $3$ or $5$ (since $14$ is not divisible by $3$ or $5$ and $b$ is not divisible by $3$ or $5$). - Therefore, $\gcd(c, 210) = 1$. - **Conclusion**: Statement I is true. 3. **Analyze Statement II**: Suppose $\gcd(c, 210) = 1$. We need to check if this implies $\gcd(a, 14) = 1$ or $\gcd(b, 15) = 1$ or both. - If $\gcd(c, 210) = 1$, then $c$ is not divisible by $2$, $3$, $5$, or $7$. - From $c = 15a + 14b$, if $c$ is not divisible by $2$ or $7$, then $a$ must not be divisible by $2$ or $7$ (since $15b$ does not affect divisibility by $2$ or $7$), hence $\gcd(a, 14) = 1$. - Similarly, if $c$ is not divisible by $3$ or $5$, then $b$ must not be divisible by $3$ or $5$ (since $14a$ does not affect divisibility by $3$ or $5$), hence $\gcd(b, 15) = 1$. - **Conclusion**: Statement II is true. 4. **Analyze Statement III**: We need to check if $\gcd(c, 210) = 1$ if and only if $\gcd(a, 14) = \gcd(b, 15) = 1$. - From the analysis in Statements I and II, we see that if either $\gcd(a, 14) = 1$ or $\gcd(b, 15) = 1$, it is sufficient for $\gcd(c, 210) = 1$. - However, Statement III claims that both $\gcd(a, 14) = 1$ and $\gcd(b, 15) = 1$ are necessary, which is not true as either one being true is sufficient. - **Conclusion**: Statement III is false. ### Final Conclusion: Statements I and II are true, but III is false. Therefore, the correct answer is $\textbf{(C)}$. CCCCC |
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