2017 AMC 12A Problems/Problem 18

Problem

Let $S(n)$ equal the sum of the digits of positive integer $n$. For example, $S(1507) = 13$. For a particular positive integer $n$, $S(n) = 1274$. Which of the following could be the value of $S(n+1)$?

$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265$

Solution 1

Note that $n\equiv S(n)\bmod 9$, so $S(n+1)-S(n)\equiv n+1-n = 1\bmod 9$. So, since $S(n)=1274\equiv 5\bmod 9$, we have that $S(n+1)\equiv 6\bmod 9$. The only one of the answer choices $\equiv 6\bmod 9$ is $\boxed{(D)=\ 1239}$.

Solution 2

One possible value of $S(n)$ would be $1275$, but this is not any of the choices. Therefore, we know that $n$ ends in $9$, and after adding $1$, the last digit $9$ carries over, turning the last digit into $0$. If the next digit is also a $9$, this process repeats until we get to a non-$9$ digit. By the end, the sum of digits would decrease by $9$ multiplied by the number of carry-overs but increase by $1$ as a result of the final carrying over. Therefore, the result must be $9x-1$ less than original value of $S(n)$, $1274$, where $x$ is a positive integer. The only choice that satisfies this condition is $\boxed{1239}$, since $(1274-1239+1) \bmod 9 = 0$. The answer is $\boxed{D}$.

Solution 3

Another way to solve this is to realize that if you continuously add the digits of the number $1274 (1 + 2 + 7 + 4 = 14, 1 + 4 = 5)$, we get $5$. Adding one to that, we get $6$. So, if we assess each option to see which one attains $6$, we would discover that $1239$ satisfies the requirement, because $1 + 2 + 3 + 9 = 15$. $1 + 5 = 6$. The answer is $\boxed{D}$.

See Also

2017 AMC 10A (ProblemsAnswer KeyResources)
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 10 Problems and Solutions
2017 AMC 12A (ProblemsAnswer KeyResources)
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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