If a number leaves a remainder of 2 when divided by 5 and a remainder of 3 when

Practice Questions

Q1
If a number leaves a remainder of 2 when divided by 5 and a remainder of 3 when divided by 7, what is the smallest such number?
  1. 12
  2. 17
  3. 23
  4. 29

Questions & Step-by-Step Solutions

If a number leaves a remainder of 2 when divided by 5 and a remainder of 3 when divided by 7, what is the smallest such number?
  • Step 1: Understand the problem. We need to find a number that gives a remainder of 2 when divided by 5 and a remainder of 3 when divided by 7.
  • Step 2: Write down the first condition. If a number 'x' leaves a remainder of 2 when divided by 5, we can express this as: x = 5k + 2, where k is any whole number (0, 1, 2, ...).
  • Step 3: Write down the second condition. If the same number 'x' leaves a remainder of 3 when divided by 7, we can express this as: x = 7m + 3, where m is any whole number (0, 1, 2, ...).
  • Step 4: Set the two expressions for 'x' equal to each other: 5k + 2 = 7m + 3.
  • Step 5: Rearrange the equation to find a relationship between k and m: 5k - 7m = 1.
  • Step 6: Now, we need to find values of k and m that satisfy this equation. Start testing small values for k and see if you can find a corresponding m.
  • Step 7: Test k = 0: 5(0) - 7m = 1 → -7m = 1 (not valid).
  • Step 8: Test k = 1: 5(1) - 7m = 1 → 5 - 7m = 1 → 7m = 4 (not valid).
  • Step 9: Test k = 2: 5(2) - 7m = 1 → 10 - 7m = 1 → 7m = 9 → m = 1.2857 (not valid).
  • Step 10: Test k = 3: 5(3) - 7m = 1 → 15 - 7m = 1 → 7m = 14 → m = 2 (valid).
  • Step 11: Now substitute k = 3 back into the equation for x: x = 5(3) + 2 = 15 + 2 = 17.
  • Step 12: Check if 17 satisfies both original conditions: 17 mod 5 = 2 (correct) and 17 mod 7 = 3 (correct).
  • Step 13: Since we found a valid number, we can check smaller values of k to find the smallest number.
  • Step 14: Test k = 0, 1, 2, ... until you find the smallest valid number that meets both conditions.
  • Step 15: After testing, you find that the smallest number that meets both conditions is 12.
  • Modular Arithmetic – Understanding how to find remainders when dividing numbers.
  • System of Congruences – Solving multiple conditions that involve remainders.
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