Difference between revisions of "2017 AMC 12A Problems/Problem 18"
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− | One possible value of <math>S(n)</math> would be <math>1275</math>, but this is not any of the choices. Ergo, we know that <math>n</math> ends in <math>9</math>, and after adding <math>1</math>, the last digit <math>9</math> carries over, turning the last digit into <math>0</math>. If the next digit is also a <math>9</math>, this process repeats until we get to a non-<math>9</math> digit. By the end, the sum of digits would decrease by <math>9</math> multiplied by the number of carry-overs but increase by 1 as a result of the final carrying over. Therefore, the result must be <math>9x-1</math> less than original value of <math>S(n)</math>, <math>1274</math>, where <math>x</math> is a positive integer. The only choice that satisfies this condition is <math>1239</math>, since <math>(1274-1239+1) \bmod 9 = 0</math>. The answer is <math>\boxed{D}</math>. | + | One possible value of <math>S(n)</math> would be <math>1275</math>, but this is not any of the choices. Ergo, we know that <math>n</math> ends in <math>9</math>, and after adding <math>1</math>, the last digit <math>9</math> carries over, turning the last digit into <math>0</math>. If the next digit is also a <math>9</math>, this process repeats until we get to a non-<math>9</math> digit. By the end, the sum of digits would decrease by <math>9</math> multiplied by the number of carry-overs but increase by <math>1</math> as a result of the final carrying over. Therefore, the result must be <math>9x-1</math> less than original value of <math>S(n)</math>, <math>1274</math>, where <math>x</math> is a positive integer. The only choice that satisfies this condition is <math>1239</math>, since <math>(1274-1239+1) \bmod 9 = 0</math>. The answer is <math>\boxed{D}</math>. |
== See Also == | == See Also == |
Revision as of 02:23, 9 February 2017
Contents
Problem
Let equal the sum of the digits of positive integer . For example, . For a particular positive integer , . Which of the following could be the value of ?
Solution
Note that , so . So, since , we have that . The only one of the answer choices is .
Solution 2
One possible value of would be , but this is not any of the choices. Ergo, we know that ends in , and after adding , the last digit carries over, turning the last digit into . If the next digit is also a , this process repeats until we get to a non- digit. By the end, the sum of digits would decrease by multiplied by the number of carry-overs but increase by as a result of the final carrying over. Therefore, the result must be less than original value of , , where is a positive integer. The only choice that satisfies this condition is , since . The answer is .
See Also
2017 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 |
2017 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 |
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