2011 AIME I Problems/Problem 11
Problem
Let be the set of all possible remainders when a number of the form , a nonnegative integer, is divided by . Let be the sum of the elements in . Find the remainder when is divided by .
Solution
Note that and . So we must find the first two integers and such that and and . Note that and will be greater than 2 since remainders of will not be possible after 2 (the numbers following will always be congruent to 0 modulo 8). Note that (see Euler's theorem) and are all distinct modulo 125 (proof below). Thus, and are the first two integers such that . All that is left is to find in mod . After some computation: To show that are distinct modulo 125, suppose for the sake of contradiction that they are not. Then, we must have at least one of or . However, writing , we can easily verify that and , giving us the needed contradiction.
See also
2011 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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