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

(Corrected numerical answer)
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==Solutions==
 
==Solutions==
 
===Solution 1===
 
===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.
+
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.  
−
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>.
+
 
 +
</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>
  
 
===Solution 2===
 
===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>.
+
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>.
 +
 
 +
===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}$.
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2013|n=II|num-b=5|num-a=7}}
 
{{AIME box|year=2013|n=II|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:01, 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$1000\N$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$\boxed{282}$.

See Also

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