Question:

Find the smallest number that leaves a remainder of 4 on division by 5, 5 on division by 6, 6 on division by 7, 7 on division by 8, and 8 on division by 9?

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For problems involving remainders, observe the pattern and check for the LCM of the divisors. This often simplifies the process of finding the smallest number that satisfies the conditions.
Updated On: Mar 10, 2025
  • 2519
  • 5039
  • 1079
  • 979
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The Correct Option is A

Solution and Explanation

The number leaves a remainder of 4 when divided by 5. This means the number can be expressed as: \[ x \equiv 4 \pmod{5} \] Similarly, we have the following conditions: \[ x \equiv 5 \pmod{6} \] \[ x \equiv 6 \pmod{7} \] \[ x \equiv 7 \pmod{8} \] \[ x \equiv 8 \pmod{9} \] These conditions suggest that the number is 1 less than a multiple of 5, 6, 7, 8, and 9. Hence, we need to find the least common multiple (LCM) of these numbers: \[ \text{LCM}(5, 6, 7, 8, 9) = 2520 \] Now, since the number must be 1 less than a multiple of the LCM, we subtract 1 from the LCM: \[ 2520 - 1 = 2519 \] Thus, the smallest number that satisfies all the conditions is 2519. Final Answer: The correct answer is (a) 2519.
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