Difference between revisions of "2013 AIME II Problems/Problem 6"
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==Solutions== | ==Solutions== | ||
===Solution 1=== | ===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. | |
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+ | </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. | ||
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+ | 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< | + | 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>. | ||
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+ | ===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}$. | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2013|n=II|num-b=5|num-a=7}} | {{AIME box|year=2013|n=II|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:01, 7 October 2019
Contents
[hide]Problem 6
Find the least positive integer such that the set of consecutive integers beginning with contains no square of an integer.
Solutions
Solution 1
The difference between consecutive integral squares must be greater than 1000. , so \implies x\geq500x=500x>500n=x-500nn^21000\Nn^2=1000n=10\sqrt{10}\sqrt{10}3^2=910=(x+3)^2=x^2+6x+9x^210=6x+9x=1/6\implies \sqrt{10}\approx 19/6$. This is 3.16.
Then,$ (Error compiling LaTeX. Unknown error_msg)n\approx 31.6n^2<1000n31531^2532^2531^2=281961532^2=283024282,000NN=282000\implies\boxed{282}x^2(x+1)^2x\ge 0(x+1)^2-x^2=2x+1x\ge 02x+1x2x+1\ge 1000x\ge \frac{999}{2}x\ge 500\overline{N000}\rightarrow \overline{N999}xx=500x^2=250000(x+1)^2=251001250000\overline{N000}310009619742x+1531^2=281961532^2=283024282000\boxed{282}$.
===Solution 3=== Let$ (Error compiling LaTeX. Unknown error_msg)xN250x500kx-500x^2(500+k)^2250000+1000k+k^22500001000k\overline{N000}\rightarrow \overline{N999}k^210001000k=31k^2=961(k+1)^2=32^2=10241000k1000x(500+31)^2=281961(500+32)^2=283024282000\boxed{282}$.
See Also
2013 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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