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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) + 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>
  
 
== See also ==
 
== See also ==

Latest revision as of 21:26, 8 October 2024

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 $n \in \lbrace0, 1, 2, 3, 4, 5, 6\rbrace$.

Now, note that $3\cdot 37 = 111$ so $30 \cdot 37 = 1110$, and $90 \cdot 37 = 3330$ so $87 \cdot 37 = 3219$. 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 $(0 + 1 + 2 + 3 + 4 + 5 + 6) + 7\cdot28 = \boxed{217}$

See also

2004 AIME I (Problems • Answer Key • Resources)
Preceded by
First Question 		Followed by
Problem 2
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All AIME Problems and Solutions

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