Find the greatest number of five digits which is divisible by $15$, $21$ and $36$.

### Concept & Strategy To find the greatest $N$-digit number divisible by a set of numbers, first find their L.C.M. Then, divide the largest possible $N$-digit number (e.g., $99999$) by this L.C.M. Subtract the resulting remainder from the $N$-digit number to get a perfect multiple. $$ \text{Required Number} = (\text{Largest } N\text{-digit number}) - \text{Remainder} $$ ### Step-by-Step Solution * **Given:** Target divisors are $15$, $21$, and $36$. We need a 5-digit max number. * The largest 5-digit number is $99999$. * Find the L.C.M. of $15$, $21$, and $36$: * $15 = 3 \times 5$ * $21 = 3 \times 7$ * $36 = 2^2 \times 3^2$ * $\text{L.C.M.} = 2^2 \times 3^2 \times 5 \times 7 = 4 \times 9 \times 35 = 1260$ * Divide $99999$ by the L.C.M. ($1260$): * $1260 \times 70 = 88200$ * $1260 \times 79 = 99540$ * Division yields a quotient of $79$ and a remainder of $459$. * Subtract the remainder from the original largest 5-digit number: * $\text{Required Number} = 99999 - 459 = 99540$ ### Exam Strategy & Shortcut Don't calculate long division if you don't have to. The correct number must be divisible by $36$, which means it must be divisible by $9$. Check the options for divisibility by $9$ (sum of digits): * $99999$ (sum $45$, divisible, but leaves rem $9$ when div by $10$/$36$ fail) * $99540$ (sum $27$, divisible by $9$) * $98260$ (sum $25$, fail) * $99640$ (sum $28$, fail) Option B is clearly the only mathematically viable distractor that perfectly handles the $9$ and $20$ constraints. ### Common Pitfall Students often confuse "greatest number" phrasing with finding the Highest Common Factor (H.C.F.). Here, "greatest number" acts as a boundary condition for the magnitude of the answer, not the factor operation. You must still use L.C.M. ### Final Answer **Therefore, the correct answer is 99540.**