2004 AIME I Problems/Problem 1
Contents
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
Solution 2
For any n we are told that it is four consecutive integers in decreasing order when read from left to right. Thus for any number() ( do not work because a digit cannot be greater than 9), is equal to or . Now we try this number for . When , and when divided by has a remainder of 28. We now notice that every time you increase by you increase by and has remainder when divided by . Thus, the remainder increases by every time you increase by . Thus,
When , the remainder equals 28
When , the remainder equals 29
When , the remainder equals 30
When , the remainder equals 31
When , the remainder equals 32
When , the remainder equals 33
When , the remainder equals 34
Thus the sum of the remainders is equal to which is equal to .
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 |
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