2013 AIME II Problems/Problem 6

Find the least positive integer $N$ such that the set of $1000$ consecutive integers beginning with $1000\cdot N$ contains no square of an integer.

Solution

Let us first observe the difference between $x^2$ and $(x+1)^2$ , for any arbitrary $x\ge 0$ . $(x+1)^2-x^2=2x+1$ . So that means for every $x\ge 0$ , the difference between that square and the next square have a difference of $2x+1$ . Now, we need to find an $x$ such that $2x+1\ge 1000$ . Solving gives $x\ge \frac{999}{2}$ , so $x\ge 500$ . Now we need to find what range of numbers has to be square-free: $\overline{N000}\rightarrow \overline{N999}$ have to all be square-free. Let us first plug in a few values of $x$ to see if we can figure anything out. $x=500$ , $x^2=250000$ , and $(x+1)^2=251001$ . Notice that this does not fit the criteria, because $250000$ is a square, whereas $\overline{N000}$ cannot be a square. This means, we must find a square, such that the last $3$ digits are close to $1000$ , but not there, such as $961$ or $974$ . Now, the best we can do is to keep on listing squares until we hit one that fits. We do not need to solve for each square: remember that the difference between consecutive squares are $2x+1$ , so all we need to do is addition. After making a list, we find that $531^2=281961$ , while $532^2=283024$ . It skipped $282000$ , so our answer is $\boxed{282}$

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2013 AIME II Problems/Problem 6

Solution