Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
m
Line 4: Line 4:
 
===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.
 
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 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 <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}}

Revision as of 15:37, 6 April 2013

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.

Solution

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 $\boxed{282}$.

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 $\boxed{282}$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_II_Problems/Problem_6&oldid=52194"