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. Ergo, 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 $\boxed{D}$ .

2017 AMC 10A (Problems • Answer Key • Resources)
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2017 AMC 12A (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

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2017 AMC 12A Problems/Problem 18

Contents

Problem

Solution 1

Solution 2

See Also