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Difference between revisions of "2004 AIME I Problems/Problem 1"

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== Solution ==
 
== Solution ==
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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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A brute-force solution to this question is fairly quick, but we'll try something slightly more clever:  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 + 28</math>.  Adding these numbers up, we get <math>0 + 1 + 2 + 3 + 4 + 5 + 6 + 7\cdot28 = 217</math>, our answer.
  
 
== See also ==
 
== See also ==

Revision as of 14:59, 22 August 2006

Problem

The digits of a positive integer $n$ are four consecutive integers in decreasing order when read from left to right. What is the sum of the possible remainders when $n$ is divided by 37?

Solution

A brute-force solution to this question is fairly quick, but we'll try something slightly more clever: our numbers have the form ${\underline{(n+3)}}\,{\underline{(n+2)}}\,{\underline{( n+1)}}\,{\underline {(n)}} = 1000(n + 3) + 100(n + 2) + 10(n + 1) + n = 3210 + 1111n$, for $n \in \{0, 1, 2, 3, 4, 5, 6\}$. Now, note that $3\cdot 37 = 111$ so $30 \cdot 37 = 1110$, and $90 \cdot 37 = 3330$ so $89 \cdot 37 = 3219$. So the remainders are all congruent to $n - 9 \pmod{37}$. However, these numbers are negative for our choices of $n$, so in fact the remainders must equal $n + 28$. Adding these numbers up, we get $0 + 1 + 2 + 3 + 4 + 5 + 6 + 7\cdot28 = 217$, our answer.

See also

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