2001 AIME I Problems/Problem 1
Contents
1
2
3
4
5
6
Problem
Solution 1
Solution 2
Solution 3
Solution 4
See also
Problem
Find the sum of all positive two-digit integers that are divisible by each of their digits.
Solution 1
Let our number be
,
. Then we have two conditions:
, or
divides into
and
divides into
and
. Thus
or
(note that if
, then
would not be a digit).
For
, we have
For
for nine possibilities, giving us a sum of
, we have
.
for four possibilities (the higher ones give
), giving us a sum of
for one possibility (again, higher ones give
), giving us a sum of
.
For
, we have
If we ignore the case
as we have been doing so far, then the sum is
.
.
Solution 2
Using casework, we can list out all of these numbers:
Solution 3
To further expand on solution 2, it would be tedious to test all
two-digit numbers. We can reduce the amount to look at by focusing
on the tens digit. First, we cannot have any number that is a multiple of
. We also note that any number with the same digits is a
number that satisfies this problem. This gives
We start from each of these numbers and constantly add the digit of the tens number of the respective number until we get a different
tens digit. For example, we look at numbers
and numbers
. This heavily reduces the numbers
we need to check, as we can deduce that any number with a tens digit of or greater that does not have two of the same digits is not a
valid number for this problem. This will give us the numbers from solution 2.
Solution 4
In this solution, we will do casework on the ones digit. Before we start, let's make some variables. Let be the ones digit, and
tens digit. Let equal our number. Our number can be expressed as
. We can easily see that
, since
, and
Therefore,
. Now, let's start with the casework.
Case 1:
Since
,
. From this, we get that
satisfies the condition.
be the
.
Case 2:
We either have
Case 3:
We have
does not divide
.
, or
. From this, we get that
. From this, we get that
and
satisfies the condition. Note that
Case 4:
We either have
or
not included for similar reasons as last time.
. From this, we get that
and
Case 5:
. From this, we get that
and
We either have
or
Continuing with this process up to
. Summing, we get that the answer is
, we get that
satisfy the condition.
was not included because
satisfy the condition.
was
satisfy the condition.
could be
. A clever way to sum would be to group the multiples of
together to get
, and then add the remaining
.
-bronzetruck2016
See also
2001 AIME I (Problems • Answer Key • Resources (http://www.artofproblemsolving.com/Forum/resources.php?
c=182&cid=45&year=2001))
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 (http://www.maa.org)'s American Mathematics
Competitions (http://amc.maa.org).
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_1&oldid=151722"
Copyright © 2021 Art of Problem Solving