Difference between revisions of "2009 AMC 12A Problems/Problem 12"
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The sum of the digits is at most <math>9+9+9=27</math>. Therefore the number is at most <math>6\cdot 27 = 162</math>. Out of the numbers <math>1</math> to <math>162</math> the one with the largest sum of digits is <math>99</math>, and the sum is <math>9+9=18</math>. Hence the sum of digits will be at most <math>18</math>. | The sum of the digits is at most <math>9+9+9=27</math>. Therefore the number is at most <math>6\cdot 27 = 162</math>. Out of the numbers <math>1</math> to <math>162</math> the one with the largest sum of digits is <math>99</math>, and the sum is <math>9+9=18</math>. Hence the sum of digits will be at most <math>18</math>. | ||
− | Also, each number with this property is divisible by <math>6</math>, therefore it is divisible by <math>3</math>, and thus also its sum of digits is divisible by <math>3</math>. | + | Also, each number with this property is divisible by <math>6</math>, therefore it is divisible by <math>3</math>, and thus also its sum of digits is divisible by <math>3</math>. Thus, the number is divisible by <math>18</math>. |
− | We only have six possibilities left for the sum of the digits: <math>3</math>, <math>6</math>, <math>9</math>, <math>12</math>, <math>15</math>, and <math>18</math> | + | We only have six possibilities left for the sum of the digits: <math>3</math>, <math>6</math>, <math>9</math>, <math>12</math>, <math>15</math>, and <math>18</math>, but since the number is divisible by <math>18</math>, the digits can only add to <math>9</math> or <math>18</math>. This leads to the integers <math>18</math>, <math>36</math>, <math>54</math>, <math>72</math>, <math>90</math>, and <math>108</math> being possibilities. We can check to see that <math>\boxed{1}</math> solution: the number <math>54</math> is the only solution that satisfies the conditions in the problem. |
=== Solution 2 === | === Solution 2 === |
Latest revision as of 19:06, 17 December 2017
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 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.