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?

• A. 2519
• B. 5039
• C. 1079
• D. 979

Hint:

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.

Answer 1

The Correct Option is A

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.