AMC 10 · 2023 · #20

Grade 9 patternnumber-theory
recursive-sequencesequences-geometricunits-digit-trackingpattern-recognition easier-related-problempattern-recognitionidentify-subproblems ↑ Prerequisites: recursive-sequencepattern-recognition
📏 Long solution 💡 4 insights 📊 Diagram
Problem
A triangular array is built row by row. The first row is a single one. Every later row starts and ends with one, and each entry between them is one plus the sum of the two entries diagonally above. The n-th row holds n numbers. Find the units digit of the sum of the numbers in row 2023.

Pick an answer.

(A)
1
(B)
3
(C)
5
(D)
7
(E)
9
How to solve
Strategy Introduce a Variable

Row 2023 has 2023 separate entries, and not one of them is wanted — only their total is. Tool #9 (Solve an Easier Related Problem) is the opening move: replace the question "what are the 2023 numbers?" with "what is their sum?", which is one number instead of 2023. Tool #4 (Introduce a Variable) then does the real work twice over. First it names that one number, S_n, so a row-to-row rule can even be written. Second, and this is the crux, it renames S_n as T_n = S_n + n, which turns a messy recurrence with a drifting +n-2 into plain doubling. Tool #16 (Change Focus) supplies the recurrence itself: instead of adding up row n entry by entry, count how many times each entry of row n-1 gets used by row n. Tool #5 (Look for a Pattern) closes the problem, since the units digits of powers of 2 repeat in a cycle of four. Tool #15 (Organize Information in More Ways) keeps a running row-sum table that every later claim can be tested against.

1STEP 1

Tabulate the row sums

Tabulate the first few sums.

n & sum of row n ; 1 & 1 ; 2 & 2 ; 3 & 5 ; 4 & 12 ; 5 & 27 ; 6 & 58
2STEP 2

Chase the sum, not the entries

Chase the sum, not the entries.

S_n = sum of the n entries in row n, S₁ = 1, S₂ = 2, S₃ = 5, S₄ = 12, S₅ = 27, S₆ = 58
3STEP 3

How often each parent is used

Interior parents are used twice.

S_n = 2_two ends + (n-2)_one +1 per interior entry + (2S_n-1 - 2)_parents, ends used once = 2S_n-1 + n - 2
4STEP 4

Shift the sequence

A small shift simplifies the rule.

T_n = S_n + n → T_n = (2S_n-1 + n - 2) + n = 2S_n-1 + 2n - 2 = 2(S_n-1 + (n-1)) = 2T_n-1
5STEP 5

Read the doubling rule

Doubling gives a power of two.

T₂ = 4 = 2², T_n = 2T_n-1 → T_n = 2ⁿ → S_n = 2ⁿ - n
6STEP 6

Use the units-digit cycle

Powers of two cycle through four units digits.

2¹ → 2, 2² → 4, 2³ → 8, 2⁴ → 6, 2⁵ → 2, … 2023 = 4(505) + 3 → 2²⁰²³ ≡ 2³ ≡ 8 (mod 10)
7STEP 7

Subtract the last digits

Subtracting gives 5.

S₂₀₂₃ = 2²⁰²³ - 2023 ≡ 8 - 3 ≡ 5 (mod 10) → (C)
8STEP 8

Check on real rows

Confirm the formula on real rows.

S₅ = 2⁵ - 5 = 27 = 1+7+11+7+1, S₆ = 2⁶ - 6 = 58 = 1+9+19+19+9+1
Answer
5
Three checks agree. (1) Size — the entries roughly double from row to row, so a row sum near 2ⁿ is expected, and 2ⁿ - n is only a small trim below it; for n = 2023 the subtracted 2023 is utterly negligible next to 2²⁰²³, yet it is exactly what shifts the last digit from 8 to 5, so the correction cannot be dropped. (2) Direct verification — the closed form was tested on rows 3 through 6, all built straight from the construction rule, and it matched every time. (3) The near-miss trap — a solver who forgets the -2023 reads off 8 and picks (D), while a solver who assumes Pascal-like pure doubling gets S_n = 2ⁿ⁻¹ and lands on a units digit of 4, which is not even a listed choice; that missing option is itself a hint that the +1 in the rule genuinely changes the answer. The value 5 from 8 - 3 is choice (C).
💡Key takeaway

When a rule builds each row from the one above, stop tracking the entries and track only the row sum — here shifting it to S_n + n makes it double every row, so S_n = 2ⁿ - n and the last digit of 2²⁰²³ - 2023 is 8 - 3 = 5.

  • Tabulate the rows and their sums
  • Chase the sum, not the entries
  • Count how often each parent is used
  • Shift the sequence to kill the drift
  • Doubling gives a power of two
  • Units digits of powers of two cycle by four
  • Subtract the last digits
  • Confirm the formula on real rows