2009 AMC 12A Problems/Problem 12
Problem
How many positive integers less than are times the sum of their digits?
Solution
Solution 1
The sum of the digits is at most . Therefore the number is at most . Out of the numbers to the one with the largest sum of digits is , and the sum is . Hence the sum of digits will be at most .
Also, each number with this property is divisible by , therefore it is divisible by , and thus also its sum of digits is divisible by . Thus, the number is divisible by .
We only have six possibilities left for the sum of the digits: , , , , , and , but since the number is divisible by , the digits can only add to or . This leads to the integers , , , , , and being possibilities. We can check to see that solution: the number is the only solution that satisfies the conditions in the problem.
Solution 2
We can write each integer between and inclusive as where and . The sum of digits of this number is , hence we get the equation . This simplifies to . Clearly for there are no solutions, hence and we get the equation . This obviously has only one valid solution , hence the only solution is the number .
Solution 3
The sum of the digits is at most . Therefore the number is at most . Since the number is times the sum of its digits, it must be divisible by , therefore also by , therefore the sum of its digits must be divisible by . With this in mind we can conclude that the number must be divisible by , not just by . Since the number is divisible by , it is also divisible by , therefore the sum of its digits is divisible by , therefore the number is divisible by , which leaves us with , and . Only is times its digits, hence the answer is .
See Also
2009 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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All AMC 12 Problems and Solutions |
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