AMC 10 · 2017 · #20

Grade 6 arithmetic
Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In
digit-summodular-arithmeticdivisibility-rules invariant-monovariantcasework ↑ Prerequisites: digit-sum
📏 Medium solution 💡 3 insights
Problem
Let S(n) be the sum of the digits of a positive integer n; for example S(1507) = 1+5+0+7 = 13. A certain n has S(n) = 1274. Among the answer choices, decide which value S(n+1) could equal.

Pick an answer.

(A)
1
(B)
3
(C)
12
(D)
1239
(E)
1265

AMC 10 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Eliminate Possibilities

We can't see n, so we work with the one thing we control: what adding 1 does to a digit sum. Tool #5 (Look for a Pattern) on small numbers reveals a clean rule — adding 1 raises the digit sum by 1, unless trailing 9s roll over to 0. Tool #4 (Introduce a Variable) names k, the count of trailing 9s, and turns the rule into a formula for S(n+1). Then, because this is multiple choice with a finite list, Tool #3 (Eliminate Possibilities) tests each choice against the formula and keeps only the one that gives a whole-number k.

1STEP 1

See what adding 1 does

S(7)→S(8) just rises by 1, but S(19)=10→S(20)=2: a trailing 9 flips to 0, losing 9 while the carry adds 1.

S(20)=S(19)-9+1=10-9+1=2
2STEP 2

Write the rule as a formula

Let k count the trailing 9s of n; adding 1 zeroes them (−9k) and bumps the next digit (+1), so from S(n)=1274 we get 1275 − 9k.

S(n+1) = 1274 - 9k + 1 = 1275 - 9k, k = 0,1,2,…
3STEP 3

Test each choice for a whole-number k

A choice c works only if 1275 − c is divisible by 9; for 1, 3, 12, 1265 the gaps 1274, 1272, 1263, 10 are not, so all four fail.

1274, 1272, 1263, 10 are not multiples of 9
4STEP 4

Confirm the survivor

1275 − 1239 = 36 = 9×4, so k=4: n ends in four 9s and its other digits sum to 1238. Such n exists, so S(n+1) can be 1239 — choice (D).

1275 - 9(4) = 1275 - 36 = 1239 → (D)
Answer
1239
Four choices were thrown out because 1275 - c left a remainder mod 9, and only 1239 survived with k=4, so the answer is unique as the problem promises. It also makes sense that S(n+1) = 1239 is less than S(n) = 1274: adding 1 here rolls four nines down to zeros, and a falling digit sum is exactly what trailing-nine carrying produces.
💡Key takeaway

Adding 1 raises a digit sum by 1, but every trailing 9 that rolls over to 0 knocks it down by 9.

  • See what adding 1 does
  • Write the rule as a formula
  • Test each choice for a whole-number k
  • Confirm the survivor

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