2013 AIME II Problems/Problem 6

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.

Solutions

Solution 1

The difference between consecutive integral squares must be greater than 1000. $(x+1)^2-x^2\geq1000$ , so $x\geq\frac{999}{2}$ \implies x\geq500 $.$ x=500 $does not work, so$ x>500 $. Let$ n=x-500 $The sum of the square of$ n $and a number a little over 1000 must result in a new perfect square. By inspection,$ n^2 $should end in a number close to but less than 1000 such that there exists$ 1000\N $within the difference of the two squares. Examine when$ n^2=1000 $. Then,$ n=10\sqrt{10} $. Estimate$ \sqrt{10} $. One example way follows.$ 3^2=9 $, so$ 10=(x+3)^2=x^2+6x+9 $.$ x^2 $is small, so$ 10=6x+9 $.$ x=1/6\implies \sqrt{10}\approx 19/6$. This is 3.16.

Then,$ (Error compiling LaTeX. Unknown error_msg)n\approx 31.6 $.$ n^2<1000 $, so$ n $could be$ 31 $. Add 500 to get the first square and 501 to get the second. Then, the two integral squares are$ 531^2 $and$ 532^2 $. Checking,$ 531^2=281961 $and$ 532^2=283024 $.$ 282,000 $straddles the two squares, which have a difference of 1063. The difference has been minimized, so$ N $is minimized$ N=282000\implies\boxed{282} $===Solution 2=== Let us first observe the difference between$ 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}$.

===Solution 3=== Let$ (Error compiling LaTeX. Unknown error_msg)x $be the number being squared. Based on the reasoning above, we know that$ N $must be at least$ 250 $, so$ x $has to be at least$ 500 $. Let$ k $be$ x-500 $. We can write$ x^2 $as$ (500+k)^2 $, or$ 250000+1000k+k^2 $. We can disregard$ 250000 $and$ 1000k $, since they won't affect the last three digits, which determines if there are any squares between$ \overline{N000}\rightarrow \overline{N999} $. So we must find a square,$ k^2 $, such that it is under$ 1000 $, but the next square is over$ 1000 $. We find that$ k=31 $gives$ k^2=961 $, and so$ (k+1)^2=32^2=1024 $. We can be sure that this skips a thousand because the$ 1000k $increments it up$ 1000 $each time. Now we can solve for$ x $:$ (500+31)^2=281961 $, while$ (500+32)^2=283024 $. We skipped$ 282000 $, so the answer is$ \boxed{282}$.

2013 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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2013 AIME II Problems/Problem 6

Contents

Problem 6

Solutions

Solution 1

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