Difference between revisions of "2004 AIME I Problems/Problem 1"
(→Solution) |
|||
| (26 intermediate revisions by 14 users not shown) | |||
| Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
| − | The digits of a positive integer <math> n </math> are four consecutive integers in decreasing order when read from left to right. What is the sum of the possible remainders when <math> n </math> is divided by 37? | + | The digits of a positive integer <math> n </math> are four consecutive integers in decreasing order when read from left to right. What is the sum of the possible remainders when <math> n </math> is divided by <math>37</math>? |
== 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)}} </math><math>= 1000(n + 3) + 100(n + 2) + 10(n + 1) + n = 3210 + 1111n</math>, for <math>n \in \lbrace0, 1, 2, 3, 4, 5, 6\rbrace</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> | ||
== See also == | == See also == | ||
| − | + | {{AIME box|year=2004|n=I|before=First Question|num-a=2}} | |
| + | |||
| + | [[Category:Intermediate Number Theory Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 21:26, 8 October 2024
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 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
