Difference between revisions of "2004 AIME I Problems/Problem 1"
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Now, note that <math>3\cdot 37 = 111</math> so <math>30 \cdot 37 = 1110</math>, and <math>90 \cdot 37 = 3330</math> so <math>87 \cdot 37 = 3219</math>. So the [[remainder]]s are all congruent to <math>n - 9 \pmod{37}</math>. However, these numbers are negative for our choices of <math>n</math>, so in fact the remainders must equal <math>n + 28</math>. | Now, note that <math>3\cdot 37 = 111</math> so <math>30 \cdot 37 = 1110</math>, and <math>90 \cdot 37 = 3330</math> so <math>87 \cdot 37 = 3219</math>. So the [[remainder]]s are all congruent to <math>n - 9 \pmod{37}</math>. However, these numbers are negative for our choices of <math>n</math>, so in fact the remainders must equal <math>n + 28</math>. | ||
| − | Adding these numbers up, we get <math>(0 + 1 + 2 + 3 + 4 + 5 + 6 | + | Adding these numbers up, we get <math>(0 + 1 + 2 + 3 + 4 + 5 + 6 + 7)\cdot28 = \boxed{217}</math>, our answer. |
== See also == | == See also == | ||
Revision as of 09:06, 7 March 2018
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
