Find the smallest number which when increased by $10$ is completely divisible by $12$, $15$, $18$, $20$ and $24$.

### Concept & Formula To find a number that becomes completely divisible by a set of given divisors only *after* a specific value $X$ is added to it, calculate the Lowest Common Multiple (L.C.M.) of the divisors and subtract $X$ from that result. $$ \text{Required Number} = \text{L.C.M.}(a, b, c, \dots) - X $$ ### Step-by-Step Solution * **Given:** The divisors are $12$, $15$, $18$, $20$, and $24$. The number must be divisible when increased by $10$. * First, find the L.C.M. of the divisors. Let's use prime factorization: * $12 = 2^2 \times 3$ * $15 = 3 \times 5$ * $18 = 2 \times 3^2$ * $20 = 2^2 \times 5$ * $24 = 2^3 \times 3$ * Identify the highest power of each prime factor present: * For $2$: $2^3 = 8$ * For $3$: $3^2 = 9$ * For $5$: $5^1 = 5$ * Multiply these maximum prime powers together: * $\text{L.C.M.} = 8 \times 9 \times 5 = 360$ * The L.C.M. ($360$) is the smallest number perfectly divisible by all. Since our target number only reaches this state *after* adding $10$, we must subtract $10$: * $\text{Required Number} = 360 - 10 = 350$ ### Exam Strategy & Shortcut Use the **Option Elimination** method combined with divisibility rules. The problem states that `Option + 10` is divisible by $18$ and $20$. * Divisibility by $20$ implies the last digit must be $0$. * Divisibility by $18$ implies it must be divisible by $9$ (sum of digits is a multiple of $9$). Test Option A: $350 + 10 = 360$. The sum of digits of $360$ is $9$, making it perfectly divisible by $9$, and it ends in $0$. It passes immediately. ### Common Pitfall A very common mistake is calculating the L.C.M. ($360$) and then intuitively adding $10$ to it because the question contains the phrase "increased by $10$". You must work backwards and *subtract* to find the original state. ### Final Answer **Therefore, the correct answer is 350.**