Difference between revisions of "2013 AIME II Problems/Problem 6"
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==Solution 4== | ==Solution 4== | ||
− | We want to find the least <math>N \in \mathbb{N}</math> such that <math>\exists m \in \mathbb{N}</math> where <math>m^2 | + | We want 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, we have, | Combining the two inequalities, we have, | ||
+ | <math>(m+1)^2 \geq m^2 + 1001</math>, | ||
− | <math> | + | <math>m \geq 500</math>. |
− | <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> | + | <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>, a contradiction. | ||
− | + | Note <math>k \geq 32</math> leads to larger <math>N</math>. | |
+ | Hence, the answer is <math>\boxed{282}</math>. | ||
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==See Also== | ==See Also== |
Revision as of 14:20, 21 July 2021
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 . The sum of the square of and a number a little over 1000 must result in a new perfect square. 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
We want to find the least such that where .
Combining the two inequalities, we have,
,
.
Let , where , then the inequalities become,
, and
For , one can verify that is the unique integer satisfying the inequalities.
For , , a contradiction.
Note leads to larger .
Hence, the answer is .
\yuxiaomatt
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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