2013 AIME II Problems/Problem 6
Find the least positive integer such that the set of consecutive integers beginning with contains no square of an integer.
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
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 .
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 .
The goal is to find the least such that where .
Combining the two inequalities leads to .
Let , where , then the inequalities become,
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 .
Very similar to 2016 AMC 12 A Problem 25: https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_25
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