2007 AIME I Problems/Problem 7
Problem
Let
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 .)
Solution
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
See also
2007 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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