Difference between revisions of "2017 AMC 12A Problems/Problem 18"
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<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265</math> | <math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265</math> | ||
| − | ==Solution== | + | ==Solution 1== |
Note that <math>n\equiv S(n)\bmod 9</math>, so <math>S(n+1)-S(n)\equiv n+1-n = 1\bmod 9</math>. So, since <math>S(n)=1274\equiv 5\bmod 9</math>, we have that <math>S(n+1)\equiv 6\bmod 9</math>. The only one of the answer choices <math>\equiv 6\bmod 9</math> is <math>\boxed{(D)=\ 1239}</math>. | Note that <math>n\equiv S(n)\bmod 9</math>, so <math>S(n+1)-S(n)\equiv n+1-n = 1\bmod 9</math>. So, since <math>S(n)=1274\equiv 5\bmod 9</math>, we have that <math>S(n+1)\equiv 6\bmod 9</math>. The only one of the answer choices <math>\equiv 6\bmod 9</math> is <math>\boxed{(D)=\ 1239}</math>. | ||
Revision as of 01:45, 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 1
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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