The least number of 4 digits which is a perfect square is

### Concept & Formula To find the least $n$-digit perfect square, identify the smallest integer whose square crosses the threshold into the $n$-digit range. The smallest 4-digit number is $1000$. We need to find the integer whose square is strictly greater than or equal to $1000$. ### Step-by-Step Solution * **Given:** We need the least 4-digit perfect square. * **Calculation / Deduction:** * The smallest 4-digit number is $1000$. * We know that $30^2 = 900$ (a 3-digit number). * We know that $40^2 = 1600$. * Let's narrow it down. We know $31^2 = 961$, which is still a 3-digit number. * The next integer is $32$. Let's calculate $32^2$. * $32^2 = (30 + 2)^2$ * Applying $(a+b)^2 = a^2 + 2ab + b^2$: $$32^2 = 900 + 120 + 4$$ $$32^2 = 1024$$ * Since $1024$ is a 4-digit number, it is the least possible 4-digit perfect square. ### Exam Strategy & Shortcut Base powers of $2$ are incredibly useful to memorize in competitive exams and computer science contexts. Knowing that $2^{10} = 1024$ instantly gives you the answer. $1024$ is famously $32^2$ ($2^5 \times 2^5 = 2^{10}$). ### Common Pitfall A common mistake is checking numbers randomly between $30$ and $40$ or relying on long division for square roots. While long division works, simply knowing basic squares around milestones (like $30^2 = 900$) and stepping up by $1$ is much faster. ### Final Answer **Therefore, the correct answer is 1024.**