The largest four-digit number which is a perfect cube, is

Concept & Logic The problem requires finding the largest perfect cube less than or equal to $9999$ (the absolute largest four-digit number). We can narrow down the search space by checking the cubes of multiples of $10$. Step-by-Step Solution * The largest four-digit number is $9999$. * Let us establish baseline cubes to find the correct range: $$20^3 = 8000$$ $$30^3 = 27000$$ * The target number is between $20$ and $30$, but much closer to $20$. * Check the cube of the next integer, $21$: $$21^3 = 21 \times 21 \times 21 = 441 \times 21 = 9261$$ * Now check the cube of $22$ to ensure we haven't missed a larger four-digit cube: $$22^3 = 22 \times 22 \times 22 = 484 \times 22 = 10648$$ * Since $10648$ is a five-digit number, the largest integer whose cube results in a four-digit number is $21$. * The corresponding perfect cube is $9261$. Exam Strategy & Shortcut Memorizing perfect cubes up to at least $25$ is highly recommended for aptitude exams. If you know that $20^3 = 8000$ and $21^3 = 9261$, you can quickly evaluate the given options. Option (a) $8000$ is too small, and option (c) $9999$ is immediately obvious as a non-cube (since the cube of a number ending in $9$ must end in $9$, but $19^3 = 6859$ and $29^3$ is near $27000$). This leaves $9261$ as the definitive answer. Common Pitfall Assuming $9999$ might be a perfect cube simply because it is the largest four-digit number overall, or selecting $8000$ because it is the most recognizable "clean" cube without checking if a larger one exists below the $10000$ threshold. Final Answer **Therefore, the correct answer is 9261.**