text: # 1992 AHSME Problems/Problem 23
  
  (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
  
  ==Problem== jj
  
  Let $S$ be a subset of $\{1,2,3,...,50\}$ such that no pair of distinct elements in $S$ has a sum divisible by $7$. What is the maximum number of elements in $S$?
  
  $\text{(A) } 6\quad \text{(B) } 7\quad \text{(C) } 14\quad \text{(D) } 22\quad \text{(E) } 23$
  
  ## Solution
  
  The fact that $x \equiv 0 \mod 7 \Rightarrow 7 \mid x$ is assumed as common knowledge in this answer.
  
  First, note that there are $8$ possible numbers that are equivalent to $1 \mod 7$, and there are $7$ possible numbers equivalent to each of $2$-$6 \mod 7$.
  
  Second, note that there can be no pairs of numbers $a$ and $b$ such that $a \equiv -b$ mod $7$, because then $a+b | 7$. These pairs are $(0,0)$, $(1,6)$, $(2,5)$, and $(3,4)$. Because $(0,0)$ is a pair, there can always be $1$ number equivalent to $0 \mod 7$, and no more.
  
  To maximize the amount of numbers in S, we will use $1$ number equivalent to $0 \mod 7$, $8$ numbers equivalent to $1$, and $14$ numbers equivalent to $2$-$5$. This is obvious if you think for a moment. Therefore the answer is $1+8+14=23$ numbers. $\fbox{E}$
explanation: The document contains a math problem from the 1992 AHSME (American High School Mathematics Examination) and its solution. The problem asks for the maximum number of elements in a subset \( S \) of \(\{1,2,3,...,50\}\) such that no pair of distinct elements in \( S \) has a sum divisible by 7. The solution is provided, and the answer is given as 23.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # 1992 AHSME Problems/Problem 23
  
  (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
  
  ==Problem== jj
  
  Let $S$ be a subset of $\{1,2,3,...,50\}$ such that no pair of distinct elements in $S$ has a sum divisible by $7$. What is the maximum number of elements in $S$?
  
  $\text{(A) } 6\quad \text{(B) } 7\quad \text{(C) } 14\quad \text{(D) } 22\quad \text{(E) } 23$
  
  ## Solution
  
  The fact that $x \equiv 0 \mod 7 \Rightarrow 7 \mid x$ is assumed as common knowledge in this answer.
  
  First, note that there are $8$ possible numbers that are equivalent to $1 \mod 7$, and there are $7$ possible numbers equivalent to each of $2$-$6 \mod 7$.
  
  Second, note that there can be no pairs of numbers $a$ and $b$ such that $a \equiv -b$ mod $7$, because then $a+b | 7$. These pairs are $(0,0)$, $(1,6)$, $(2,5)$, and $(3,4)$. Because $(0,0)$ is a pair, there can always be $1$ number equivalent to $0 \mod 7$, and no more.
  
  To maximize the amount of numbers in S, we will use $1$ number equivalent to $0 \mod 7$, $8$ numbers equivalent to $1$, and $14$ numbers equivalent to $2$-$5$. This is obvious if you think for a moment. Therefore the answer is $1+8+14=23$ numbers. $\fbox{E}$
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math problem from the 1992 AHSME (American High School Mathematics Examination) and its solution. The problem asks for the maximum number of elements in a subset \( S \) of \(\{1,2,3,...,50\}\) such that no pair of distinct elements in \( S \) has a sum divisible by 7. The solution is provided, and the answer is given as 23.
  
  Label: question-and-answer