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

Revision as of 21:02, 7 October 2019

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 $1000\N$ (Error compiling LaTeX. Unknown error_msg) within the difference of the two squares. Examine when $n^2=1000$ . Then, $n=10\sqrt{10}$ . One example way to estimate $\sqrt{10}$ follows.

, so .  is small, so . . This is 3.16.

Then, $n\approx 31.6$ . $n^2<1000$ , 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 $N=282000\implies\boxed{282}$

~BJHHar

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 $x$ 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 4: / Line 4: @@
 ==Solutions==
 ===Solution 1===
-The difference between consecutive integral squares must be greater than 1000. <math>(x+1)^2-x^2\geq1000</math>, so <math>x\geq\frac{999}{2}</math>\implies x\geq500<math>. </math>x=500<math> does not work, so </math>x>500<math>. Let </math>n=x-500<math> The sum of the square of </math>n<math> and a number a little over 1000 must result in a new perfect square. By inspection,  </math>n^2<math> should end in a number close to but less than 1000 such that there exists </math>1000\N<math> within the difference of the two squares. Examine when </math>n^2=1000<math>. Then, </math>n=10\sqrt{10}<math>. Estimate </math>\sqrt{10}<math>. One example way follows.
+The difference between consecutive integral squares must be greater than 1000. <math>(x+1)^2-x^2\geq1000</math>, so <math>x\geq\frac{999}{2}\implies x\geq500</math>. <math>x=500</math> does not work, so <math>x>500</math>. Let <math>n=x-500</math> The sum of the square of <math>n</math> and a number a little over 1000 must result in a new perfect square. By inspection,  <math>n^2</math> should end in a number close to but less than 1000 such that there exists <math>1000\N</math> within the difference of the two squares. Examine when <math>n^2=1000</math>. Then, <math>n=10\sqrt{10}</math>. One example way to estimate <math>\sqrt{10}</math> follows.
-  </math>3^2=9<math>, so </math>10=(x+3)^2=x^2+6x+9<math>. </math>x^2<math> is small, so </math>10=6x+9<math>. </math>x=1/6\implies \sqrt{10}\approx 19/6<math>. This is 3.16.
+  <math>3^2=9</math>, so <math>10=(x+3)^2=x^2+6x+9</math>. <math>x^2</math> is small, so <math>10=6x+9</math>. <math>x=1/6\implies \sqrt{10}\approx 19/6</math>. This is 3.16.
-Then, </math>n\approx 31.6<math>. </math>n^2<1000<math>, so </math>n<math> could be </math>31<math>. Add 500 to get the first square and 501 to get the second. Then, the two integral squares are </math>531^2<math> and </math>532^2<math>. Checking, </math>531^2=281961<math> and </math>532^2=283024<math>. </math>282,000<math> straddles the two squares, which have a difference of 1063. The difference has been minimized, so </math>N<math> is minimized </math>N=282000\implies\boxed{282}<math>
+Then, <math>n\approx 31.6</math>. <math>n^2<1000</math>, so <math>n</math> could be <math>31</math>. Add 500 to get the first square and 501 to get the second. Then, the two integral squares are <math>531^2</math> and <math>532^2</math>. Checking, <math>531^2=281961</math> and <math>532^2=283024</math>. <math>282,000</math> straddles the two squares, which have a difference of 1063. The difference has been minimized, so <math>N</math> is minimized <math>N=282000\implies\boxed{282}</math>
+~BJHHar
 ===Solution 2===
-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 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{282}</math>.
 ===Solution 3===
-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}$.
+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>.
 ==See Also==
 {{AIME box|year=2013|n=II|num-b=5|num-a=7}}
 {{MAA Notice}}

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