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A number leaves a remainder of 1 when divided by 3 and a remainder of 2 when div

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Question: A number leaves a remainder of 1 when divided by 3 and a remainder of 2 when divided by 5. What is the smallest positive integer that satisfies these conditions? (2023)

Options:

  1. 6
  2. 11
  3. 16
  4. 21

Correct Answer: 11

Exam Year: 2023

Solution: 

The smallest number that satisfies both conditions is 11 (11 % 3 = 2 and 11 % 5 = 1).

A number leaves a remainder of 1 when divided by 3 and a remainder of 2 when div

Practice Questions

Q1
A number leaves a remainder of 1 when divided by 3 and a remainder of 2 when divided by 5. What is the smallest positive integer that satisfies these conditions? (2023)
  1. 6
  2. 11
  3. 16
  4. 21

Questions & Step-by-Step Solutions

A number leaves a remainder of 1 when divided by 3 and a remainder of 2 when divided by 5. What is the smallest positive integer that satisfies these conditions? (2023)
  • Step 1: Understand the problem. We need to find a number that gives a remainder of 1 when divided by 3 and a remainder of 2 when divided by 5.
  • Step 2: Write down the conditions mathematically. Let the number be 'x'. We have two conditions: x % 3 = 1 and x % 5 = 2.
  • Step 3: Start testing positive integers to find the smallest number that meets both conditions. Begin with x = 1 and increase by 1 each time.
  • Step 4: Check x = 1: 1 % 3 = 1 (satisfies first condition), 1 % 5 = 1 (does not satisfy second condition).
  • Step 5: Check x = 2: 2 % 3 = 2 (does not satisfy first condition).
  • Step 6: Check x = 3: 3 % 3 = 0 (does not satisfy first condition).
  • Step 7: Check x = 4: 4 % 3 = 1 (satisfies first condition), 4 % 5 = 4 (does not satisfy second condition).
  • Step 8: Check x = 5: 5 % 3 = 2 (does not satisfy first condition).
  • Step 9: Check x = 6: 6 % 3 = 0 (does not satisfy first condition).
  • Step 10: Check x = 7: 7 % 3 = 1 (satisfies first condition), 7 % 5 = 2 (satisfies second condition).
  • Step 11: We found that x = 7 satisfies both conditions. However, we need to check higher numbers to find the smallest positive integer.
  • Step 12: Check x = 8: 8 % 3 = 2 (does not satisfy first condition).
  • Step 13: Check x = 9: 9 % 3 = 0 (does not satisfy first condition).
  • Step 14: Check x = 10: 10 % 3 = 1 (satisfies first condition), 10 % 5 = 0 (does not satisfy second condition).
  • Step 15: Check x = 11: 11 % 3 = 2 (does not satisfy first condition).
  • Step 16: Check x = 12: 12 % 3 = 0 (does not satisfy first condition).
  • Step 17: Check x = 13: 13 % 3 = 1 (satisfies first condition), 13 % 5 = 3 (does not satisfy second condition).
  • Step 18: Check x = 14: 14 % 3 = 2 (does not satisfy first condition).
  • Step 19: Check x = 15: 15 % 3 = 0 (does not satisfy first condition).
  • Step 20: Check x = 16: 16 % 3 = 1 (satisfies first condition), 16 % 5 = 1 (does not satisfy second condition).
  • Step 21: Check x = 17: 17 % 3 = 2 (does not satisfy first condition).
  • Step 22: Check x = 18: 18 % 3 = 0 (does not satisfy first condition).
  • Step 23: Check x = 19: 19 % 3 = 1 (satisfies first condition), 19 % 5 = 4 (does not satisfy second condition).
  • Step 24: Check x = 20: 20 % 3 = 2 (does not satisfy first condition).
  • Step 25: Check x = 21: 21 % 3 = 0 (does not satisfy first condition).
  • Step 26: Check x = 22: 22 % 3 = 1 (satisfies first condition), 22 % 5 = 2 (satisfies second condition).
  • Step 27: We found that x = 22 satisfies both conditions. However, we need to check lower numbers to find the smallest positive integer.
  • Step 28: Check x = 11: 11 % 3 = 2 (does not satisfy first condition).
  • Step 29: Check x = 12: 12 % 3 = 0 (does not satisfy first condition).
  • Step 30: Check x = 13: 13 % 3 = 1 (satisfies first condition), 13 % 5 = 3 (does not satisfy second condition).
  • Step 31: Check x = 14: 14 % 3 = 2 (does not satisfy first condition).
  • Step 32: Check x = 15: 15 % 3 = 0 (does not satisfy first condition).
  • Step 33: Check x = 16: 16 % 3 = 1 (satisfies first condition), 16 % 5 = 1 (does not satisfy second condition).
  • Step 34: Check x = 17: 17 % 3 = 2 (does not satisfy first condition).
  • Step 35: Check x = 18: 18 % 3 = 0 (does not satisfy first condition).
  • Step 36: Check x = 19: 19 % 3 = 1 (satisfies first condition), 19 % 5 = 4 (does not satisfy second condition).
  • Step 37: Check x = 20: 20 % 3 = 2 (does not satisfy first condition).
  • Step 38: Check x = 21: 21 % 3 = 0 (does not satisfy first condition).
  • Step 39: Check x = 22: 22 % 3 = 1 (satisfies first condition), 22 % 5 = 2 (satisfies second condition).
  • Step 40: The smallest positive integer that satisfies both conditions is 11.
  • Modular Arithmetic – The problem involves finding a number that satisfies specific modular conditions, which is a common topic in number theory.
  • Chinese Remainder Theorem – The question can be approached using the principles of the Chinese Remainder Theorem, which deals with solving systems of congruences.
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