Competition · AMC preparation · step 4 of 4
AMC 8 · 2023 · #16
Grade 4 patterncounting
Pick an answer.
AMC 8 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Count the letters in row 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.
Reading off a repeating PQR cycle in 20 slots is exactly what Grade 4 'generate a number or shape pattern from a rule' calls for.
4.OA.C.5Look For A PatternCount the letters in row 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.
A shifted version of the same rule still produces a pattern — just the starting letter changes.
4.OA.C.5Look For A PatternCount the letters in row 3
Row 3 is shifted by two (R, P, Q …); six cycles plus the leftover R, P give 7 Ps, 6 Qs, 7 Rs.
Same rule, shifted again — Grade 4 pattern generation handles all three row types.
4.OA.C.5Look For A PatternAdd the three rows
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.
Solving the easier 3 × 20 block first and seeing 20-20-20 is a Grade 3 'identify the arithmetic pattern' move.
3.OA.D.9Solve An Easier Related ProblemRepeat the block six times
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.
Multiplying 6 × 20 = 120 is a Grade 3 multiplication fact done three times.
Across rows 1 through 18 each of the three letters P, Q, R appears exactly 120 times, so those rows add the same amount to every letter's total.
▸ Why?
Rows 1 to 18 break into six identical three-row blocks, because the whole arrangement repeats itself every three rows down the grid.
▸ Why?
Each row is the cycle P, Q, R shifted by its row number, and shifting by three steps is one full turn of that length-3 cycle, so row 4 lands back on row 1's letters, row 5 on row 2's, and every later row copies the one three rows above it.
▸ Why?
One three-row block holds 20 of every letter, because its 20 columns each show P, Q, and R exactly once.
▸ Why?
Reading down a single column the letter steps one place along the P, Q, R cycle each row, so three stacked rows land on three different positions of the cycle and cover all three letters before any repeat.
▸ Why?
Six identical blocks that each hold 20 of a letter give 120 of that letter, since six equal groups of 20 is 6 × 20.
Handle the leftover rows
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.
Splitting off the two leftover rows and adding their counts is a Grade 4 multi-step word-problem move.
4.OA.A.3Identify SubproblemsAdd the two pieces
Combine: 120, 120, 120 plus 13, 14, 13 gives 133 Ps, 134 Qs, 133 Rs — choice (C).
Recombining the easy pieces gives the final answer — the Grade 4 multi-step problem finish line.
4.OA.A.3Identify SubproblemsThis AMC 8 problem only needs Grade 4 shape-and-number patterns you already know — find the smallest repeating block, count it, and multiply!
- Count the letters in row 1
- Count the letters in row 2
- Count the letters in row 3
- Add the three rows
- Repeat the block six times
- Handle the leftover rows
- Add the two pieces
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