2013 AIME II Problems/Problem 6
Contents
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
See Also
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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