Difference between revisions of "2001 AIME I Problems/Problem 1"
m (→Solution 2) |
|||
Line 13: | Line 13: | ||
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> | 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> | ||
+ | |||
+ | == Solution 3 == | ||
+ | |||
+ | To further expand on solution 2, it would be tedious to test all <math>90</math> 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 <math>10</math>. We also note that any number with the same digits is a number that satisfies this problem. This gives <cmath>11, 22, 33, ... 99.</cmath> We start from each of these numbers and constantly add the digit of the tens number of the respective number. For example, we look at numbers <math>11, 12, 13, ... 19</math> and numbers <math>22, 24, 26, 28</math>. This heavily reduces the numbers we need to check, as we can deduce that any number with a tens digit of <math>5</math> 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. | ||
== See also == | == See also == |
Revision as of 20:03, 7 March 2021
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 .
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:
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 We start from each of these numbers and constantly add the digit of the tens number of the respective number. 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.
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.