Difference between revisions of "2013 AIME II Problems/Problem 6"
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+ | ==Problem 6== | ||
Find the least positive integer <math>N</math> such that the set of <math>1000</math> consecutive integers beginning with <math>1000\cdot N</math> contains no square of an integer. | Find the least positive integer <math>N</math> such that the set of <math>1000</math> consecutive integers beginning with <math>1000\cdot N</math> contains no square of an integer. | ||
− | ==Solution== | + | ==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}\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. | ||
+ | |||
+ | 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}</math>. | ||
+ | |||
+ | ===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>. | ||
+ | |||
+ | |||
+ | Combining the two inequalities leads to <math>(m+1)^2 \geq m^2 + 1001, m \geq 500</math>. | ||
+ | |||
+ | |||
+ | Let <math>m = k + 500</math>, where <math>k \in \mathbb{W}</math>, then the inequalities become, | ||
+ | |||
+ | <math>N \geq \frac{(k+500)^2 + 1}{1000} = \frac{k^2 + 1}{1000} + k + 250</math>, and | ||
+ | |||
+ | <math>N \leq \frac{(k+501)^2}{1000} - 1 = \frac{(k+1)^2}{1000} + k + 250.</math> | ||
+ | |||
+ | |||
+ | For <math>k=31</math>, one can verify that <math>N = 282</math> is the unique integer satisfying the inequalities. | ||
+ | |||
+ | For <math>k \leq 30</math>, <math>k + 250 < \frac{k^2 + 1}{1000} + k + 250 \leq N</math> <math>\leq \frac{(k+1)^2}{1000} + k + 250 \leq \frac{(30+1)^2}{1000} + k + 250 < k + 251</math>, | ||
+ | |||
+ | i.e., <math>k + 250 < N < k + 251</math>, a contradiction. | ||
+ | |||
+ | Note <math>k \geq 32</math> leads to larger <math>N</math>(s). | ||
+ | |||
+ | Hence, the answer is <math>\boxed{282}</math>. | ||
+ | |||
+ | ~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 | ||
+ | {{AIME box|year=2013|n=II|num-b=5|num-a=7}} | ||
+ | {{MAA Notice}} |
Revision as of 18:42, 15 June 2024
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 . does not work, so . Let . By inspection, should end in a number close to but less than 1000 such that there exists within the difference of the two squares. Examine when . Then, . One example way to estimate follows.
, so . is small, so . . This is 3.16.
Then, . , so could be . Add 500 to get the first square and 501 to get the second. Then, the two integral squares are and . Checking, and . straddles the two squares, which have a difference of 1063. The difference has been minimized, so is minimized
~BJHHar
Solution 2
Let us first observe the difference between and , for any arbitrary . . So that means for every , the difference between that square and the next square have a difference of . Now, we need to find an such that . Solving gives , so . Now we need to find what range of numbers has to be square-free: have to all be square-free. Let us first plug in a few values of to see if we can figure anything out. , , and . Notice that this does not fit the criteria, because is a square, whereas cannot be a square. This means, we must find a square, such that the last digits are close to , but not there, such as or . 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 , so all we need to do is addition. After making a list, we find that , while . It skipped , so our answer is .
Solution 3
Let be the number being squared. Based on the reasoning above, we know that must be at least , so has to be at least . Let be . We can write as , or . We can disregard and , since they won't affect the last three digits, which determines if there are any squares between . So we must find a square, , such that it is under , but the next square is over . We find that gives , and so . We can be sure that this skips a thousand because the increments it up each time. Now we can solve for : , while . We skipped , so the answer is .
Solution 4
The goal is to find the least such that where .
Combining the two inequalities leads to .
Let , where , then the inequalities become,
, and
For , one can verify that is the unique integer satisfying the inequalities.
For , ,
i.e., , a contradiction.
Note leads to larger (s).
Hence, the answer is .
~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) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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All AIME Problems and Solutions |
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