Problem:
Let be the set of all perfect squares whose rightmost three digits in base are . Let be the set of all numbers of the form , where is in . In other words, is the set of numbers that result when the last three digits of each number in are truncated. Find the remainder when the tenth smallest element of is divided by .
Solution:
Let be an element of . Then for some integer and . Because cannot divide both and must divide one of them. Also, because is a multiple of must be a multiple of . It follows that the elements of are precisely the numbers of the form . The smallest such number is when . The tenth smallest element of is therefore , and the corresponding element of is . Hence the desired remainder is .
The problems on this page are the property of the MAA's American Mathematics Competitions