Difference between revisions of "2001 AIME I Problems/Problem 1"
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== Solution 2 == | == Solution 2 == | ||
| − | Using casework, we can list out all of these numbers: < | + | Using casework, we can list out all of these numbers: <cmath>11+12+15+22+24+33+36+44+48+55+66+77+88+99=\boxed{630}</cmath>. |
== See also == | == See also == | ||
Revision as of 22:06, 14 January 2021
Contents
[hide]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/a/4/da4c711275941be127b69f80cb22d70119da8d04.png)
.
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. 
