2012 AIME I Problems/Problem 10
Problem 10
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
It is apparent that for a perfect square to satisfy the constraints, we must have or Now in order for to be a multiple of at least one of and must be a multiple of and since and are in different residue classes mod one term must have all the factors of and thus must be a multiple of Furthermore, each of and must have at least two factors of since otherwise could not possibly be divisible by So therefore the conditions are satisfied if either or is divisible by or equivalently if Counting up from to we see that the tenth value of is and thus the corresponding element in is
See also
2012 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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