Difference between revisions of "2004 AIME I Problems/Problem 1"
(→Solution 2) |
Williamschen (talk | contribs) (→Solution 2) |
||
| (6 intermediate revisions by 2 users not shown) | |||
| Line 2: | Line 2: | ||
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>? | 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 1== |
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>. | ||
| Line 9: | Line 9: | ||
Adding these numbers up, we get <math>(0 + 1 + 2 + 3 + 4 + 5 + 6) + 7\cdot28 = \boxed{217}</math> | Adding these numbers up, we get <math>(0 + 1 + 2 + 3 + 4 + 5 + 6) + 7\cdot28 = \boxed{217}</math> | ||
| − | + | == Solution 2== | |
| + | There are only <math>7</math> possible values of <math>n</math>: <math>3210, 4321, 5432, \cdots, 9876</math>. | ||
| + | |||
| + | <math>3210</math> gives a remainder of <math>28</math> when divided by <math>37</math>. To calculate the remainders of the other integers, notice that each number is <math>1111</math> more than the previous number. Since <math>1111</math> gives a remainder of <math>1</math> when divided by <math>37</math>, the remainders of the other integers will be <math>29, 30, 31, \cdots, 34</math>. | ||
| + | |||
| + | Therefore, our answer is <math>28 + 29 + 30 + 31 + 32 + 33 + 34 = \frac{28 + 34}{2} \cdot 7 = \boxed{217}</math>. | ||
| + | ~Viliciri | ||
== See also == | == See also == | ||
Latest revision as of 22:56, 2 December 2024
Contents
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 1
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 
Solution 2
There are only 
possible values of : 
.
gives a remainder of 
when divided by . To calculate the remainders of the other integers, notice that each number is 
more than the previous number. Since gives a remainder of 
when divided by , the remainders of the other integers will be 
.
Therefore, our answer is 
.
~Viliciri
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. 
