2007 AIME I Problems/Problem 7
Find the remainder when is divided by 1000. ( is the greatest integer less than or equal to , and is the least integer greater than or equal to .)
The ceiling of a number minus the floor of a number is either equal to zero (if the number is an integer); otherwise, it is equal to 1. Thus, we need to find when or not is an integer.
The change of base formula shows that . For the term to cancel out, is a power of . Thus, is equal to the sum of all the numbers from 1 to 1000, excluding all powers of 2 from to .
The formula for the sum of an arithmetic sequence and the sum of a geometric sequence yields that our answer is .
Simplifying, we get The answer is
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