The greatest number of four digits which is divisible by 15, 25, 40 and 75 is

### Concept & Formula To find the greatest $n$-digit number divisible by a set of numbers, first find the Least Common Multiple (LCM) of those numbers. Then, find the largest multiple of this LCM that does not exceed the maximum possible $n$-digit value. * Greatest 4-digit number = $9999$ * Required Number = $9999 - (\text{Remainder of } \frac{9999}{\text{LCM}})$ ### Step-by-Step Solution **Given:** * Divisors: $15, 25, 40, 75$ * Target: Greatest 4-digit number **Calculation:** * Step 1: Find the LCM of $15, 25, 40,$ and $75$. * $15 = 3 \times 5$ * $25 = 5^2$ * $40 = 2^3 \times 5$ * $75 = 3 \times 5^2$ $$\text{LCM} = 2^3 \times 3 \times 5^2 = 8 \times 3 \times 25 = 600$$ * Step 2: Divide the greatest 4-digit number ($9999$) by the LCM ($600$) to find the remainder. $$9999 \div 600 = 16 \text{ with a remainder of } 399$$ * Step 3: Subtract this remainder from $9999$ to get the largest exact multiple. $$\text{Required Number} = 9999 - 399 = 9600$$ ### Exam Strategy & Shortcut Instead of dividing $9999$ by $600$, check the options directly for divisibility by the LCM ($600$). For a number to be divisible by $600$, it must end in at least two zeros and its remaining leading digits must be divisible by $6$. * Look at the options starting from the largest: * (d) $9800 \rightarrow 98$ is not divisible by $6$. * (c) $9600 \rightarrow 96 \div 6 = 16$. This works perfectly! Since $9600$ is the largest choice that satisfies this condition, it is the answer. ### Common Pitfall Students sometimes accidentally add the difference $(600 - 399 = 201)$ to $9999$, arriving at $10200$. While this is a multiple of $600$, it is a 5-digit number, failing the explicit 4-digit requirement. ### Final Answer **Therefore, the correct answer is 9600.**