Find the greatest number of five digits which is exactly divisible by 47.

### Concept & Logic To find the greatest $N$-digit number divisible by a given divisor, divide the largest mathematically possible $N$-digit number by the divisor. The remainder represents the excess amount. Subtract this remainder from the original number to find the greatest divisible number. ### Step-by-Step Solution - **Given:** We need the largest 5-digit number divisible by 47. - **Calculation:** The absolute largest 5-digit number is 99999. - Divide 99999 by 47 to find the remainder. - $99999 = 47 \times 2127 + 30$. - The remainder is 30. This means 99999 is 30 units greater than a perfect multiple of 47. - Subtract the remainder from the base number: $99999 - 30 = 99969$. ### Exam Strategy & Shortcut Start with 99999 and perform standard long division. The moment you find the final remainder (30), immediately subtract it from 99999. Never attempt to add the difference (divisor - remainder) for "greatest $N$-digit" questions, as adding even 1 to 99999 will flip it into a 6-digit number. ### Common Pitfall A common mistake is adding $(47 - 30)$ to 99999, which gives 100016. While 100016 is perfectly divisible by 47, it violates the constraint of being a 5-digit number. ### Final Answer Therefore, the correct answer is **99969**.