Question:
The least number which when divided by 5, 6, 7 and 8 leaves a remainder 3 but when divided by 9 leaves no remainder, is
The least number which when divided by 5, 6, 7 and 8 leaves a remainder 3 but when divided by 9 leaves no remainder, is
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Number = LCM × k + common remainder. Then test k values for the additional condition.
Updated On: Apr 20, 2026
- 1677
- 1683
- 2523
- 3363
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Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Find LCM of divisors, then find number that fits remainder and divisibility conditions.
Step 2: Detailed Explanation:
LCM of 5, 6, 7, 8 = LCM(5,6,7,8).
5 = 5, 6 = 2×3, 7 = 7, 8 = 2³ → LCM = 2³ × 3 × 5 × 7 = 8 × 3 × 5 × 7 = 840.
Number = 840k + 3 (leaves remainder 3).
Must be divisible by 9. Check k=1: 843 ÷ 9 = 93.66 (no). k=2: 1683 ÷ 9 = 187 exactly. So number = 1683.
Step 3: Final Answer:
1683
Find LCM of divisors, then find number that fits remainder and divisibility conditions.
Step 2: Detailed Explanation:
LCM of 5, 6, 7, 8 = LCM(5,6,7,8).
5 = 5, 6 = 2×3, 7 = 7, 8 = 2³ → LCM = 2³ × 3 × 5 × 7 = 8 × 3 × 5 × 7 = 840.
Number = 840k + 3 (leaves remainder 3).
Must be divisible by 9. Check k=1: 843 ÷ 9 = 93.66 (no). k=2: 1683 ÷ 9 = 187 exactly. So number = 1683.
Step 3: Final Answer:
1683
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