Question

The greatest 4-digit number which when divided by $20,24$ and $45$ leaves a remainder of $11$ in each case is

• A. $9999$
• B. $9998$
• C. $9997$
• D. $9731$

Answer 1

The correct option is D $9731$

The LCM of $20,24$ and $45$ is $360$.
The greatest four-digit number $=9999$.
Divide $9999$ by $360$. The remainder is $279$.
The greatest four-digit number divisible by $360$ is $9999-279=9720$.
Now, the greatest four-digit number divisible by $360$ and leaving a remainder $11$ is $9720+11=9731$.