2004 AIME I Problems/Problem 1
Problem
The digits of a positive integer are four consecutive integers in decreasing order when read from left to right. What is the sum of the possible remainders when is divided by ?
Solution
A brute-force solution to this question is fairly quick, but we'll try something slightly more clever: our numbers have the form , for .
Now, note that so , and so . So the remainders are all congruent to . However, these numbers are negative for our choices of , so in fact the remainders must equal .
Adding these numbers up, we get , our answer.
See also
2004 AIME I (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.