text: ## % Change
  
  From the 2013 South Africa Olympiad:
  
  A is a 2-digit number and B is a 3-digit number such that A increased by B% equals B reduced by A%. Find all possible pairs (A, B).
  
  The information presented yields the equation
  
  $A(1+\frac{B}{100}) = B(1-\frac{A}{100})$.
  
  This leads to
  
  $A(100+B) = B(100 - A)$,
  
  so that
  
  $100A+2AB = 100B$,
  
  and hence
  
  $A = \frac{50B}{50+B}$.
  
  Since A is an integer, it follows that 50+B divides 50B. Well, 50+B also divides 50(50+B) = 2500+50B, so in fact 50+B must divide 2500. Remembering that B is a 3-digit number yields just three possible values of B: 200, 450 and 575. Substituting back into the equation above yields corresponding values for A of 40, 45 and 46 respectively. As a final check, we do indeed have
  
  40 increased by 200% = 200 reduced by 40% = 120,
  
  45 increased by 450% = 450 reduced by 45% = 247.5,
  
  46 increased by 575% = 575 reduced by 46% = 310.5.
  
  Therefore the set of possible pairs (A, B) is (40, 200), (45, 450), (46, 575).
  
  A beautiful problem, borne out of a natural question that anyone fluent in percentages might ask themselves.
  
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explanation: The document presents a math problem from the 2013 South Africa Olympiad. It describes a problem involving two numbers, A and B, and their relationship through percentages. The document provides a detailed solution, including the derivation of the equation and the final set of possible pairs (A, B) that satisfy the given conditions. The solution is verified with specific examples.
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: ## % Change
  
  From the 2013 South Africa Olympiad:
  
  A is a 2-digit number and B is a 3-digit number such that A increased by B% equals B reduced by A%. Find all possible pairs (A, B).
  
  The information presented yields the equation
  
  $A(1+\frac{B}{100}) = B(1-\frac{A}{100})$.
  
  This leads to
  
  $A(100+B) = B(100 - A)$,
  
  so that
  
  $100A+2AB = 100B$,
  
  and hence
  
  $A = \frac{50B}{50+B}$.
  
  Since A is an integer, it follows that 50+B divides 50B. Well, 50+B also divides 50(50+B) = 2500+50B, so in fact 50+B must divide 2500. Remembering that B is a 3-digit number yields just three possible values of B: 200, 450 and 575. Substituting back into the equation above yields corresponding values for A of 40, 45 and 46 respectively. As a final check, we do indeed have
  
  40 increased by 200% = 200 reduced by 40% = 120,
  
  45 increased by 450% = 450 reduced by 45% = 247.5,
  
  46 increased by 575% = 575 reduced by 46% = 310.5.
  
  Therefore the set of possible pairs (A, B) is (40, 200), (45, 450), (46, 575).
  
  A beautiful problem, borne out of a natural question that anyone fluent in percentages might ask themselves.
  
  Advertisements
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document presents a math problem from the 2013 South Africa Olympiad. It describes a problem involving two numbers, A and B, and their relationship through percentages. The document provides a detailed solution, including the derivation of the equation and the final set of possible pairs (A, B) that satisfy the given conditions. The solution is verified with specific examples.
  
  Label: question-and-answer