Difference between revisions of "2001 AIME I Problems/Problem 1"
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== Solution 2 == | == Solution 2 == | ||
| − | + | By testing every 2-digit number, 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> | |
== Solution 3 == | == Solution 3 == | ||
Revision as of 19:32, 3 December 2024
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
By testing every 2-digit number, 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
In this solution, we will do casework on the ones digit.
Before we start, let's make some variables. Let be the ones digit, and be the tens digit. Let 
equal our number. Our number can be expressed as 
. We can easily see that 
, since 
, and 
. Therefore, 
.
Now, let's start with the casework.
Case 1: 
Since , 
. From this, we get that 
satisfies the condition.
Case 2: 
We either have , or 
. From this, we get that 
and 
satisfy the condition.
Case 3: 
We have 
. From this, we get that 
satisfies the condition. Note that was not included because 
does not divide 
.
Case 4: 
We either have or 
. From this, we get that 
and 
satisfy the condition. was not included for similar reasons as last time.
Case 5: 
We either have or 
. From this, we get that 
and 
satisfy the condition.
Continuing with this process up to 
, we get that could be 
. Summing, we get that the answer is 
. A clever way to sum would be to group the multiples of 
together to get 
, and then add the remaining 
.
-bronzetruck2016
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. 
