Difference between revisions of "2004 AIME I Problems/Problem 1"
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[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
Revision as of 11:24, 27 April 2008
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 | ||
