Difference between revisions of "2013 AIME II Problems/Problem 6"

Revision as of 18:00, 25 November 2017

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

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 $\boxed{261}$ .

Solution 2

Let $x$ 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 $\boxed{282}$ .

@@ Line 5: / Line 5: @@
 ===Solution 1===
 Let us first observe the difference between <math>x^2</math> and <math>(x+1)^2</math>, for any arbitrary <math>x\ge 0</math>. <math>(x+1)^2-x^2=2x+1</math>. So that means for every <math>x\ge 0</math>, the difference between that square and the next square have a difference of <math>2x+1</math>. Now, we need to find an <math>x</math> such that <math>2x+1\ge 1000</math>. Solving gives <math>x\ge \frac{999}{2}</math>, so <math>x\ge 500</math>. Now we need to find what range of numbers has to be square-free: <math>\overline{N000}\rightarrow \overline{N999}</math> have to all be square-free.
-Let us first plug in a few values of <math>x</math> to see if we can figure anything out. <math>x=500</math>, <math>x^2=250000</math>, and <math>(x+1)^2=251001</math>. Notice that this does not fit the criteria, because <math>250000</math> is a square, whereas <math>\overline{N000}</math> cannot be a square. This means, we must find a square, such that the last <math>3</math> digits are close to <math>1000</math>, but not there, such as <math>961</math> or <math>974</math>. 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 <math>2x+1</math>, so all we need to do is addition. After making a list, we find that <math>531^2=281961</math>, while <math>532^2=283024</math>. It skipped <math>282000</math>, so our answer is <math>\boxed{282}</math>.
+Let us first plug in a few values of <math>x</math> to see if we can figure anything out. <math>x=500</math>, <math>x^2=250000</math>, and <math>(x+1)^2=251001</math>. Notice that this does not fit the criteria, because <math>250000</math> is a square, whereas <math>\overline{N000}</math> cannot be a square. This means, we must find a square, such that the last <math>3</math> digits are close to <math>1000</math>, but not there, such as <math>961</math> or <math>974</math>. 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 <math>2x+1</math>, so all we need to do is addition. After making a list, we find that <math>531^2=281961</math>, while <math>532^2=283024</math>. It skipped <math>282000</math>, so our answer is <math>\boxed{261}</math>.
 ===Solution 2===
 Let <math>x</math> be the number being squared. Based on the reasoning above, we know that <math>N</math> must be at least <math>250</math>, so <math>x</math> has to be at least <math>500</math>. Let <math>k</math> be <math>x-500</math>. We can write <math>x^2</math> as <math>(500+k)^2</math>, or <math>250000+1000k+k^2</math>. We can disregard <math>250000</math> and <math>1000k</math>, since they won't affect the last three digits, which determines if there are any squares between <math>\overline{N000}\rightarrow \overline{N999}</math>. So we must find a square, <math>k^2</math>, such that it is under <math>1000</math>, but the next square is over <math>1000</math>. We find that <math>k=31</math> gives <math>k^2=961</math>, and so <math>(k+1)^2=32^2=1024</math>. We can be sure that this skips a thousand because the <math>1000k</math> increments it up <math>1000</math> each time. Now we can solve for <math>x</math>: <math>(500+31)^2=281961</math>, while <math>(500+32)^2=283024</math>. We skipped <math>282000</math>, so the answer is <math>\boxed{282}</math>.

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