Difference between revisions of "2004 AIME I Problems/Problem 1"
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== Solution == | == Solution == | ||
| − | {{ | + | A brute-force solution to this question is fairly quick, but we'll try something more clever none the less: our numbers have the form <math>{\underline{(n+3)}}\,{\underline{(n+2)}}\,{\underline{( n+1)}}\,{\underline {(n)}} = 1000(n + 3) + 100(n + 2) + 10(n + 1) + n = 3210 + 1111n</math>, for <math>n \in \{0, 1, 2, 3, 4, 5, 6\}</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>89 \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 + 26</math>. Adding these numbers up, we get <math>0 + 1 + 2 + 3 + 4 + 5 + 6 + 7\cdot26 = 203</math>, our answer. |
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== See also == | == See also == | ||
* [[2004 AIME I Problems/Problem 2| Next problem]] | * [[2004 AIME I Problems/Problem 2| Next problem]] | ||
* [[2004 AIME I Problems]] | * [[2004 AIME I Problems]] | ||
Revision as of 14:57, 22 August 2006
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 37?
Solution
A brute-force solution to this question is fairly quick, but we'll try something more clever none the less: 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.
