Difference between revisions of "2017 AMC 12A Problems/Problem 18"
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==Solution 2== | ==Solution 2== | ||
| − | 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 01:23, 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
