2001 AIME I Problems/Problem 1
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: 
and 
, or 
divides into 
and divides into 
. Thus 
or 
(note that if 
, then would not be a digit).
- For
, we have 
for nine possibilities, giving us a sum of 
. - For
, we have 
for four possibilities (the higher ones give 
), giving us a sum of 
. - For
, we have 
for one possibility (again, higher ones give ), giving us a sum of 
.
, we have 
for nine possibilities, giving us a sum of 
.
, we have 
for four possibilities (the higher ones give 
), giving us a sum of 
.
, we have 
for one possibility (again, higher ones give 
.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: ![\[11+12+15+22+24+33+36+44+48+55+66+77+88+99=\boxed{630}.\]](http://latex.artofproblemsolving.com/d/d/c/ddc68df2b734514c416f9a91b9bb57b991b8c9de.png)
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 ![\[11, 22, 33, ... 99.\]](http://latex.artofproblemsolving.com/c/1/3/c13f4d21f276d81252757727a49f65f39c2ae390.png)
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
See also
| 2001 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 | ||
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