Difference between revisions of "2001 AIME I Problems/Problem 1"
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We either have <math>b=1</math> or <math>b=5</math>. From this, we get that <math>n=15</math> and <math>n=55</math> satisfy the condition. | We either have <math>b=1</math> or <math>b=5</math>. From this, we get that <math>n=15</math> and <math>n=55</math> satisfy the condition. | ||
− | Continuing with this process up to <math>a=9</math>, we get that <math>n</math> could be <math>11, 12, 22, 33, 24, 44, 15, 55, 36, 66, 77, 48, 88, 99</math>. Summing, we get that the answer is <math>\boxed{630}</math>. A clever way to sum would be to group the multiples of <math>11</math> together to get <math>11+22+\dots+99=(45)(11)=495</math> and then add the remaining <math>12+24+15+36+48=135</math>. | + | Continuing with this process up to <math>a=9</math>, we get that <math>n</math> could be <math>11, 12, 22, 33, 24, 44, 15, 55, 36, 66, 77, 48, 88, 99</math>. Summing, we get that the answer is <math>\boxed{630}</math>. A clever way to sum would be to group the multiples of <math>11</math> together to get <math>11+22+\dots+99=(45)(11)=495</math>, and then add the remaining <math>12+24+15+36+48=135</math>. |
<i>-bronzetruck2016<i> | <i>-bronzetruck2016<i> |
Latest revision as of 17:14, 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
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.