Difference between revisions of "2004 AIME I Problems/Problem 1"
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Adding these numbers up, we get <math>(0 + 1 + 2 + 3 + 4 + 5 + 6) + 7\cdot28 = \boxed{217}</math>, our answer. | Adding these numbers up, we get <math>(0 + 1 + 2 + 3 + 4 + 5 + 6) + 7\cdot28 = \boxed{217}</math>, our answer. | ||
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| + | ==Solution 2 == | ||
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| + | 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 0-9, n is equal to <math>(d)+10(d+1)+100(d+2)+1000(d+3)</math> or <math>1111d +3210</math> | ||
== See also == | == See also == | ||
Revision as of 22:09, 21 July 2018
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 
, our answer.
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 0-9, n is equal to 
or 
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. 
