Difference between revisions of "2004 AIME I Problems/Problem 1"
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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)}} </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>. | 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> | + | 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>, our answer. |
Revision as of 17:52, 25 March 2011
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 , our answer.
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 |