AMC 8 · 2023 · #16
Grade 4 patterncountingProblem
The letters P,Q, and R are entered into a 20×20 table according to the pattern shown below. How many Ps, Qs, and Rs will appear in the completed table?
Pick an answer.
AMC 8 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A $20 \times 20$ grid is filled with the letters $P, Q, R$ following the shown pattern: row $1$ begins $P, Q, R, P, Q, R, \dots$; row $2$ begins $Q, R, P, Q, R, P, \dots$ (shifted by one); row $3$ begins $R, P, Q, R, P, Q, \dots$ (shifted by two); then the row pattern repeats. Count the total number of $P$s, $Q$s, and $R$s once the whole $20 \times 20$ grid is filled.
Givens: Grid size: $20 \times 20 = 400$ cells; Three letters $P, Q, R$ placed on a diagonal pattern that shifts by one each row; The same $3$-row block repeats down the grid (rows $1,2,3$ then rows $4,5,6$, $\dots$); Each row contains $20$ cells; Answer choices: (A) $132$ Ps, $134$ Qs, $134$ Rs; (B) $133, 133, 134$; (C) $133, 134, 133$; (D) $134, 132, 134$; (E) $134, 133, 133$
Unknowns: How many $P$s, how many $Q$s, and how many $R$s appear in the completed $20 \times 20$ grid
Understand
Restated: A $20 \times 20$ grid is filled with the letters $P, Q, R$ following the shown pattern: row $1$ begins $P, Q, R, P, Q, R, \dots$; row $2$ begins $Q, R, P, Q, R, P, \dots$ (shifted by one); row $3$ begins $R, P, Q, R, P, Q, \dots$ (shifted by two); then the row pattern repeats. Count the total number of $P$s, $Q$s, and $R$s once the whole $20 \times 20$ grid is filled.
Givens: Grid size: $20 \times 20 = 400$ cells; Three letters $P, Q, R$ placed on a diagonal pattern that shifts by one each row; The same $3$-row block repeats down the grid (rows $1,2,3$ then rows $4,5,6$, $\dots$); Each row contains $20$ cells; Answer choices: (A) $132$ Ps, $134$ Qs, $134$ Rs; (B) $133, 133, 134$; (C) $133, 134, 133$; (D) $134, 132, 134$; (E) $134, 133, 133$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #9 Solve an Easier Related Problem, #7 Identify Subproblems
The grid is built from a tiny repeating unit — three rows ($P\,Q\,R\,\dots$ then $Q\,R\,P\,\dots$ then $R\,P\,Q\,\dots$) 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 $20 \times 20$. Tool #9 (Easier Related Problem) lets us first solve the very clean $18 \times 3$ piece (six whole blocks of three rows, no leftovers), and Tool #7 (Identify Subproblems) splits the $20 \times 20$ 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.
Execute — Answer: C
4.OA.C.5 Step 1 - Count the letters in row $1$ by reading off the pattern $P, Q, R, P, Q, R, \dots$ for $20$ columns.
- The cycle $PQR$ uses $3$ columns and contains one of each letter.
- Since $20 = 6 \cdot 3 + 2$, six full cycles fill columns $1\!-\!18$ (giving $6$ of each letter) and the leftover columns $19, 20$ are the next two letters in the cycle: $P, Q$.
💡 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.5 Step 2 - Row $2$ is the same cycle but shifted by one, so it starts $Q, R, P, Q, R, P, \dots$.
- Six full $QRP$ cycles fill columns $1\!-\!18$ ($6$ of each letter); the leftover columns $19, 20$ are the next two: $Q, R$.
💡 A shifted version of the same rule still produces a pattern — just the starting letter changes.
4.OA.C.5 Step 3 - Row $3$ is shifted by two, so it starts $R, P, Q, R, P, Q, \dots$.
- Six full $RPQ$ cycles in columns $1\!-\!18$ give $6$ of each letter; the leftover columns $19, 20$ are $R, P$.
💡 Same rule, shifted again — Grade 4 pattern generation handles all three row types.
3.OA.D.9 Step 4 - Add rows $1, 2, 3$ together to find the totals for one full $3 \times 20$ block.
- Notice that within the block each letter appears the same number of times, because the block is the smallest repeating unit of the pattern.
💡 Solving the easier $3 \times 20$ block first and seeing $20$-$20$-$20$ is a Grade 3 'identify the arithmetic pattern' move.
3.OA.C.7 Step 5 - Row $4$ uses the same rule as row $1$ (shift by three is no shift), so the $3$-row counts $(7, 7, 6), (6, 7, 7), (7, 6, 7)$ repeat every $3$ rows.
- Since $18 = 6 \cdot 3$, rows $1\!-\!18$ contain exactly $6$ copies of the $3 \times 20$ block.
💡 Multiplying $6 \times 20 = 120$ is a Grade 3 multiplication fact done three times.
4.OA.A.3 Step 6 - Handle the leftover rows $19$ and $20$.
- Row $19$ behaves like row $1$ (since $19 = 6 \cdot 3 + 1$), giving $7$ P, $7$ Q, $6$ R.
- Row $20$ behaves like row $2$ (since $20 = 6 \cdot 3 + 2$), giving $6$ P, $7$ Q, $7$ R.
💡 Splitting off the two leftover rows and adding their counts is a Grade 4 multi-step word-problem move.
4.OA.A.3 Step 7 - Add the two pieces together: rows $1\!-\!18$ contribute $120, 120, 120$ and rows $19\!-\!20$ contribute $13, 14, 13$.
- The totals are $133$ Ps, $134$ Qs, $133$ Rs, which matches choice $(C)$.
💡 Recombining the easy pieces gives the final answer — the Grade 4 multi-step problem finish line.
4.OA.C.5 Count the letters in row $1$ by reading off the pattern $P, Q, R, P, Q, R, \dots 4.OA.C.5 Row $2$ is the same cycle but shifted by one, so it starts $Q, R, P, Q, R, P, \d 4.OA.C.5 Row $3$ is shifted by two, so it starts $R, P, Q, R, P, Q, \dots$. Six full $RPQ 3.OA.D.9 Add rows $1, 2, 3$ together to find the totals for one full $3 \times 20$ block. 3.OA.C.7 Row $4$ uses the same rule as row $1$ (shift by three is no shift), so the $3$-r 4.OA.A.3 Handle the leftover rows $19$ and $20$. Row $19$ behaves like row $1$ (since $19 4.OA.A.3 Add the two pieces together: rows $1\!-\!18$ contribute $120, 120, 120$ and rows Review
Reasonableness: Sanity check the sum: $133 + 134 + 133 = 400$, which equals $20 \times 20$ — every cell is accounted for. Also, $400 \div 3 = 133$ remainder $1$, so the three counts must be very close to $400/3 \approx 133.3$, 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).
Alternative: Use Tool #6 (Guess and Check) on the answer choices: the three numbers must add to $400$. Choices (A) $132+134+134 = 400$, (B) $133+133+134 = 400$, (C) $133+134+133 = 400$, (D) $134+132+134 = 400$, (E) $134+133+133 = 400$ — all five sum to $400$, so this filter alone is not enough. But it tells us that two letters appear $133$ times and one appears $134$ times in (B), (C), (E). A quick look at the diagram shows that in the visible $5 \times 5$ sub-grid the counts are $P: 8, Q: 9, R: 8$ — $Q$ is the most common — pointing to (C).
CCSS standards used (min grade 4)
3.OA.C.7Fluently multiply and divide within 100 (Computing $6 \times 20 = 120$ for the row counts across six full $3$-row blocks.)3.OA.D.9Identify arithmetic patterns and explain using properties of operations (Recognizing that the $3 \times 20$ block contains the same number of each letter ($20, 20, 20$) because it is the smallest repeating unit.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Combining the rows $1\!-\!18$ subtotal with the rows $19\!-\!20$ subtotal to get the final $133, 134, 133$.)4.OA.C.5Generate a number or shape pattern following a given rule (Reading off the repeating $PQR$ cycle across $20$ columns of each row type to count letters in rows $1, 2, 3$.)
⭐ This AMC 8 problem only needs Grade 4 shape-and-number patterns you already know — find the smallest repeating block, count it, and multiply!
⭐ This AMC 8 problem only needs Grade 4 shape-and-number patterns you already know — find the smallest repeating block, count it, and multiply!
More like this
Same archetype — closest grade level first.
