Find the least number which when divided by $6$, $7$, $8$, $9$ and $12$ leaves the same remainder $1$ in each case.

### Concept & Formula To find the least number that leaves a constant remainder $R$ when divided by multiple numbers, you first find the lowest common multiple (L.C.M.) of those divisors to get a perfectly divisible base, and then add the remainder $R$. $$ \text{Required Number} = \text{L.C.M.}(a, b, c, \dots) + R $$ ### Step-by-Step Solution * **Given:** The divisors are $6$, $7$, $8$, $9$, and $12$. The constant remainder is $1$. * First, find the L.C.M. of $6$, $7$, $8$, $9$, and $12$. Let's prime factorize: * $6 = 2 \times 3$ * $7 = 7$ (Prime) * $8 = 2^3$ * $9 = 3^2$ * $12 = 2^2 \times 3$ * Take the highest power of each prime factor: * For $2$: $2^3 = 8$ * For $3$: $3^2 = 9$ * For $7$: $7^1 = 7$ * Multiply them to get the L.C.M.: * $\text{L.C.M.} = 8 \times 9 \times 7 = 72 \times 7 = 504$ * This means $504$ is exactly divisible by all those numbers. * Now, add the required remainder ($1$) to this L.C.M.: * $\text{Required Number} = 504 + 1 = 505$ ### Exam Strategy & Shortcut Use **Divisibility Rules on the Options**. The correct answer must leave a remainder of $1$ when divided by $9$. Subtract $1$ from the options and check if the result is divisible by $9$: * $A: 504 - 1 = 503$ (Sum is $8$, not div by $9$) * $B: 505 - 1 = 504$ (Sum is $9$, **Divisible**) * $C: 503 - 1 = 502$ (Sum is $7$, not div by $9$) * $D: 512 - 1 = 511$ (Sum is $7$, not div by $9$) Option B is the only viable candidate in seconds! ### Common Pitfall A very frequent mistake is adding the remainder $1$ to the numbers *before* taking the L.C.M., or forgetting to add the remainder entirely at the end (choosing $504$). Always calculate the pure L.C.M. first, and only append the remainder at the final step. ### Final Answer **Therefore, the correct answer is 505.**