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}\implies x\geq500</math>. <math>x=500</math> does not work, so <math>x>500</math>. Let <math>n=x-500</math> | + | 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>. | ||
− | ==Solution 4== | + | ===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 | ||
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+ | ==Video Solution== | ||
+ | https://youtu.be/Rjx-0hAfQ6E?si=sr0N7dWeMg1jH5Bq | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
==See Also== | ==See Also== |
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) | ||
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Followed by Problem 7 | |
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