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

### Concept & Logic Number format: $N = \text{LCM}(5, 6, 7, 8)k + 3$. $N$ must also be a multiple of $9$. ### Step-by-Step Solution * **Find LCM:** $\text{LCM}(5, 6, 7, 8) = 840$. * **Expression:** $N = 840k + 3$. * **Condition:** $N$ is divisible by $9$. $840k + 3 = (837k + 3k) + 3$. Since $837$ is divisible by $9$, we need $(3k + 3)$ to be divisible by $9$. If $k=1$, $6$ (No). If $k=2$, $9$ (Yes). * **Calculate $N$:** $840(2) + 3 = 1683$. ### Exam Strategy & Shortcut Check divisibility by $9$ (sum of digits = multiple of $9$): (a) $1+6+7+7 = 21$ (No) (b) $1+6+8+3 = 18$ (Yes) (c) $2+5+2+3 = 12$ (No) (d) $3+3+6+3 = 15$ (No) ### Common Pitfall Assuming the LCM itself satisfies the second condition. Always verify the secondary divisibility condition. ### Final Answer **Therefore, the correct answer is 1683.**