2017 AMC 10A Problems/Problem 20

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 $n \equiv S(n) \pmod{9}$ . This can be seen from the fact that $\sum_{k=0}^{n}10^{k}a_k \equiv \sum_{k=0}^{n}a_k \pmod{9}$ . Thus, if $S(n) = 1274$ , then $n \equiv 5 \pmod{9}$ , and thus $n+1 \equiv S(n+1) \equiv 6 \pmod{9}$ . The only answer choice that is $6 \pmod{9}$ is $\boxed{\textbf{(D)}\ 1239}$ .

Solution 2

We can find out that the least number of digits the number $N$ is $142$ , with $141$ $9$ 's and one $5$ . By randomly mixing the digits up, we are likely to get: $9999$ ... $9995999$ ... $9999$ . By adding 1 to this number, we get: $9999$ ... $9996000$ ... $0000$ . Knowing that this number is ONLY divisible by $9$ when $6$ is subtracted, we can subtract 6 from every available choice, and see if the number is divisible by $9$ afterwards. After subtracting 6 from every number, we can conclude that $1233$ (originally $1239$ ) is the only number divisible by 9. So our answer is $\boxed{\textbf{(D)}\ 1239}$ .

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 10A Problems/Problem 20

Contents

Problem

Solution 1

Solution 2

See Also