AMC 8 · 2023 · #16

Grade 4 patterncounting

Archetype Lattice / Grid Pattern from Small Cases └ Sub-archetype Periodic Placement Formula

pattern-recognitionmodular-arithmeticmulti-digit-arithmetic pattern-recognitionidentify-subproblems ↑ Prerequisites: pattern-recognitionmultiples

📏 Long solution 💡 3 insights 📊 Diagram

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Problem

A 20 × 20 grid is filled with the letters P, Q, R following the shown pattern: row 1 begins P, Q, R, P, Q, R, …; row 2 begins Q, R, P, Q, R, P, … (shifted by one); row 3 begins R, P, Q, R, P, Q, … (shifted by two); then the row pattern repeats. Count the total number of Ps, Qs, and Rs once the whole 20 × 20 grid is filled.

Pick an answer.

(A)

132 Ps, 134 Qs, 134 Rs

(B)

133 Ps, 133 Qs, 134 Rs

(C)

133 Ps, 134 Qs, 133 Rs

(D)

134 Ps, 132 Qs, 134 Rs

(E)

134 Ps, 133 Qs, 133 Rs

AMC 8 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Look for a Pattern

The grid is built from a tiny repeating unit — three rows (P Q R … then Q R P … then R P Q …) that repeats forever downward. Tool #5 (Look for a Pattern) is built for exactly this: study the smallest repeating block, count letters inside it, then multiply by how many copies of the block fit into 20 × 20. Tool #9 (Easier Related Problem) lets us first solve the very clean 18 × 3 piece (six whole blocks of three rows, no leftovers), and Tool #7 (Identify Subproblems) splits the 20 × 20 grid into 'six full 3-row blocks (rows 1 - 18)' plus 'two leftover rows (rows 19, 20)' — two easy pieces instead of one hard one.

1STEP 1

Read row 1's PQR cycle across 20 columns: 20 = 6·3 + 2, so six full cycles plus the leftover P, Q give 7 Ps, 7 Qs, 6 Rs.

Row 1: 6 + 1 = 7 Ps, 6 + 1 = 7 Qs, 6 Rs

2STEP 2

Row 2 is the cycle shifted by one (Q, R, P …); six cycles plus the leftover Q, R give 6 Ps, 7 Qs, 7 Rs.

Row 2: 6 Ps, 6 + 1 = 7 Qs, 6 + 1 = 7 Rs

3STEP 3

Row 3 is shifted by two (R, P, Q …); six cycles plus the leftover R, P give 7 Ps, 6 Qs, 7 Rs.

Row 3: 6 + 1 = 7 Ps, 6 Qs, 6 + 1 = 7 Rs

4STEP 4

Add rows 1-3, the smallest repeating block: 7+6+7, 7+7+6, 6+7+7 give 20 Ps, 20 Qs, 20 Rs — a perfectly even block.

P: 7+6+7 = 20, Q: 7+7+6 = 20, R: 6+7+7 = 20

5STEP 5

Row 4 repeats row 1, so the block recurs every 3 rows; 18 = 6·3, so rows 1-18 hold 6 blocks: 120 Ps, 120 Qs, 120 Rs.

Rows 1 - 18: 6 × 20 = 120 Ps, 120 Qs, 120 Rs

6STEP 6

Rows 19, 20 act like rows 1, 2: (7,7,6) + (6,7,7) gives 13 Ps, 14 Qs, 13 Rs from the two leftover rows.

Rows 19, 20: (7+6) Ps = 13, (7+7) Qs = 14, (6+7) Rs = 13

7STEP 7

Combine: 120, 120, 120 plus 13, 14, 13 gives 133 Ps, 134 Qs, 133 Rs — choice (C).

P: 120 + 13 = 133, Q: 120 + 14 = 134, R: 120 + 13 = 133 → (C)

Answer

133 Ps, 134 Qs, 133 Rs

Sanity check the sum: 133 + 134 + 133 = 400, which equals 20 × 20 — every cell is accounted for. Also, 400 ÷ 3 = 133 remainder 1, so the three counts must be very close to

\frac{400}{3}

≈ 133.3, with exactly one of them being 134 and the other two 133. Only (B), (C), and (E) satisfy this sum check; (A) and (D) fail by giving 400 but with the wrong distribution shape. The diagram-derived pattern picks out Q as the 'extra' letter, confirming (C).

💡Key takeaway

This AMC 8 problem only needs Grade 4 shape-and-number patterns you already know — find the smallest repeating block, count it, and multiply!

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