Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
m (break up a long formula)
Line 3: Line 3:
  
 
== Solution ==
 
== Solution ==
−
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.
+
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)}} </math><math>= 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 12:48, 11 March 2007

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems/Problem_1&oldid=13852"