2017 AMC 10A Problem 20

Suppose $n$ ends in exactly $k$ nines ($k \ge 0$). Adding $1$ turns those $k$ nines into zeros and increases the next digit by $1$, so
$$S(n+1) = S(n) + 1 - 9k = 1275 - 9k.$$
Hence $S(n+1) \equiv 1275 \pmod 9$. The digit sum of $1275$ is $15$, so $1275 \equiv 6 \pmod 9$, and $S(n+1)$ must leave remainder $6$ when divided by $9$.

Check the choices by their own digit sums: $1 \to 1$; $3 \to 3$; $12 \to 3$; $1239 \to 15 \to 6$; $1265 \to 14 \to 5$. Only $1239$ is $6 \pmod 9$.

It is attainable: $1275 - 1239 = 36 = 9 \cdot 4$, so take $n$ ending in exactly four nines with the remaining digits summing to $1274 - 36 = 1238$ (for example $6$ followed by $154$ eights, then $9999$). Then $S(n+1) = 1239$.

The answer is $\boxed{\textbf{(D)}\ 1239}$.

A digit sum is congruent to the number itself mod $9$, so $S(n+1) \equiv n + 1 \equiv S(n) + 1 \pmod 9$ no matter how many carries occur; carries only subtract multiples of $9$. Divisibility-by-9 invariants like this one are the fastest way to eliminate answer choices in digit problems.

Skip the carry analysis: since $S(m) \equiv m \pmod 9$ for every $m$, $S(n+1) \equiv n + 1 \equiv S(n) + 1 = 1275 \equiv 6 \pmod 9$. Test the five choices mod $9$; only $1239$ survives.

• Assuming S(n+1) = S(n) + 1 = 1275 (not a choice) and then guessing, without accounting for carries.
• Picking $1265$ because $1275 - 1265 = 10$ "looks like a carry," when each carry removes exactly $9$.