AMC 10 · 2002 · #21

Grade 6 pattern
recursive-sequenceunits-digit-trackingpattern-recognition systematic-enumeration ↑ Prerequisites: recursive-sequenceunits-digit-tracking
📏 Long solution 💡 3 insights
Problem
A sequence begins 4,7, and from the third term on each term is the units digit of the sum of the two terms just before it. Let S_n be the sum of the first n terms. Find the smallest n for which S_n exceeds 10,000.

Pick an answer.

(A)
1992
(B)
1999
(C)
2001
(D)
2002
(E)
2004
How to solve
Strategy Look for a Pattern

Adding two thousand terms by hand is out of the question, and the answer choices sit too close together for estimation alone to separate them. Tool #2 (Make a Systematic List) writes out terms until something repeats. Tool #5 (Look for a Pattern) turns that repetition into a proof rather than a hope: since every term is decided by the two before it, the moment a consecutive pair reappears the entire sequence repeats forever. Tool #9 (Solve an Easier Related Problem) then replaces summing thousands of terms with summing one short block and multiplying. Only the last stretch is left, where the running total crosses 10,000 partway through a block, and that is short enough for tool #6 (Guess and Check), stepping one term at a time.

1STEP 1

Write terms until a pair repeats

Writing terms by the rule, the opening pair 4,7 returns at positions 13 and 14, so the block is 12 long.

4,7,1,8,9,7,6,3,9,2,1,3 | 4,7,…
2STEP 2

Prove the block repeats forever

Equal pairs force identical futures, so the sequence is periodic from the very first term.

(a₁₃,a₁₄)=(a₁,a₂) → a_n+12=a_n for all n ≥ 1
3STEP 3

Sum one block, then many blocks

One block sums to 60, so k whole blocks give 60k over 12k terms.

4+7+1+8+9+7+6+3+9+2+1+3=60, S₁₂k=60k
4STEP 4

Find the last full block below target

Since 10,000 = 60 · 166 + 40, the last full block ends at S₁₉₉₂ = 9960, still short.

10,000=60 · 166+40, S₁₉₉₂=9960, S₂₀₀₄=10,020
5STEP 5

Walk the next block term by term

Stepping into the next block, S₁₉₉₈ = 9996 and S₁₉₉₉ = 10,002, so the answer is 1999, choice (B).

S₁₉₉₈=9996 ≤ 10,000 < 10,002=S₁₉₉₉ → (B)
Answer
1999
A rough estimate lands in the right neighbourhood: the block averages 60/12=5 per term, so reaching 10,000 should take about 10,000/5=2000 terms, and 1999 is right there. The answer choices are a menu of near-misses around the true crossing. Choice (A) 1992 is the last full-block mark, where the total is only 9960 and the target has not been reached at all. Choice (E) 2004 is the next full-block mark, where the total is 10,020 — over the target, but late, since the crossing already happened five terms in. Choices (C) 2001 and (D) 2002 also sit past the crossing, so their totals do exceed 10,000 but they are not the smallest such n. The two exact endpoint totals settle it: S₁₉₉₈=9996 fails the strict inequality and S₁₉₉₉=10,002 passes.
💡Key takeaway

When a rule only looks at the last two terms, a repeated pair means the whole sequence loops forever — so add up one loop, multiply, and then walk the leftover terms by hand.

  • Write terms until a pair repeats
  • Prove the block repeats forever
  • Sum one block, then many blocks
  • Find the last full block below target
  • Walk the next block term by term