2001 AIME I Problems/Problem 1
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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.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.