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

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==Solutions==
 
==Solutions==
 
===Solution 1===
 
===Solution 1===
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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>1000N</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.  
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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>. By inspection,  <math>n^2</math> should end in a number close to but less than 1000 such that there exists <math>1000N</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.
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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>.
  
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==Solution 4==
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===Solution 4===
 
The goal is to find the least <math>N \in \mathbb{N}</math> such that <math>\exists m \in \mathbb{N}</math> where <math>m^2 + 1 \leq 1000N, 1000N + 1000 \leq (m+1)^2</math>.
 
The goal is to find the least <math>N \in \mathbb{N}</math> such that <math>\exists m \in \mathbb{N}</math> where <math>m^2 + 1 \leq 1000N, 1000N + 1000 \leq (m+1)^2</math>.
  
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~yuxiaomatt
 
~yuxiaomatt
 +
 +
 +
==Video Solution==
 +
https://youtu.be/Rjx-0hAfQ6E?si=sr0N7dWeMg1jH5Bq
 +
 +
~MathProblemSolvingSkills.com
 +
  
 
==See Also==
 
==See Also==

Latest revision as of 18:42, 15 June 2024

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 $n=x-500$. By inspection, $n^2$ should end in a number close to but less than 1000 such that there exists $1000N$ 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. 

$3^2=9$, 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, $n\approx 31.6$. $n^2<1000$, 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 $\boxed{282}$.

Solution 4

The goal is to find the least $N \in \mathbb{N}$ such that $\exists m \in \mathbb{N}$ where $m^2 + 1 \leq 1000N, 1000N + 1000 \leq (m+1)^2$.


Combining the two inequalities leads to $(m+1)^2 \geq m^2 + 1001, m \geq 500$.


Let $m = k + 500$, where $k \in \mathbb{W}$, then the inequalities become,

$N \geq \frac{(k+500)^2 + 1}{1000} = \frac{k^2 + 1}{1000} + k + 250$, and

$N \leq \frac{(k+501)^2}{1000} - 1 = \frac{(k+1)^2}{1000} + k + 250.$


For $k=31$, one can verify that $N = 282$ is the unique integer satisfying the inequalities.

For $k \leq 30$, $k + 250 < \frac{k^2 + 1}{1000} + k + 250 \leq N$ $\leq \frac{(k+1)^2}{1000} + k + 250 \leq \frac{(30+1)^2}{1000} + k + 250 < k + 251$,

i.e., $k + 250 < N < k + 251$, a contradiction.

Note $k \geq 32$ leads to larger $N$(s).

Hence, the answer is $\boxed{282}$.

~yuxiaomatt


Video Solution

https://youtu.be/Rjx-0hAfQ6E?si=sr0N7dWeMg1jH5Bq

~MathProblemSolvingSkills.com


See Also

Very similar to 2016 AMC 12 A Problem 25: https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_25

2013 AIME II (Problems • Answer Key • Resources)
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