Difference between revisions of "2001 AIME I Problems/Problem 1"
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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. | 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 until we get a different tens digit. 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. | 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 until we get a different tens digit. 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. | ||
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+ | == Solution 4 == | ||
== See also == | == See also == |
Revision as of 15:40, 16 April 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 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
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.