Difference between revisions of "2001 AIME I Problems/Problem 1"
Line 16: | Line 16: | ||
[[Category:Introductory Number Theory Problems]] | [[Category:Introductory Number Theory Problems]] | ||
− | |||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:33, 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
.
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.