Difference between revisions of "2001 AIME I Problems/Problem 1"
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Revision as of 19:32, 4 July 2013
Problem
Find the sum of all positive two-digit integers that are divisible by each of their digits.
Solution
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 
.
Using casework, we can list out all of these numbers: 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
