Difference between revisions of "2004 AIME I Problems/Problem 1"
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Thus the sum of the remainders is equal to <math>28+29+30+31+32+33+34</math> which is equal to <math>217</math>. | Thus the sum of the remainders is equal to <math>28+29+30+31+32+33+34</math> which is equal to <math>217</math>. | ||
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-David Camacho | -David Camacho |
Revision as of 10:29, 26 January 2019
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
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.
Solution 2
For any n we are told that it is four consecutive integers in decreasing order when read from left to right. Thus for any number()
(
do not work because a digit cannot be greater than 9),
is equal to
or
. Now we try this number for
. When
,
and
when divided by
has a remainder of 28. We now notice that every time you increase
by
you increase
by
and
has remainder
when divided by
. Thus, the remainder increases by
every time you increase
by
. Thus,
When , the remainder equals 28
When , the remainder equals 29
When , the remainder equals 30
When , the remainder equals 31
When , the remainder equals 32
When , the remainder equals 33
When , the remainder equals 34
Thus the sum of the remainders is equal to which is equal to
.
-David Camacho
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.