AMC 8 · 2023 · #16

Easy mode Grade 4

Problem

Picture a $20\times 20$ table — that's $20$ rows and $20$ columns, $400$ boxes in all.

We fill the boxes with only three letters: $\text{P}$ , $\text{Q}$ , and $\text{R}$ . The picture below shows the rule for how the letters get placed.

If you look down the first column, the letters repeat in the order P, Q, R, P, Q, R, P, Q, R, $\dots$ — the same three letters in the same loop, over and over.

The next column shifts the loop by one, starting Q, R, P, Q, R, P, $\dots$

The column after that shifts again, starting R, P, Q, R, P, Q, $\dots$

Then the pattern of starting letters loops back, and the fourth column starts with P again, just like the first.

After every one of the $400$ boxes is filled this way, count how many $\text{P}$ s, how many $\text{Q}$ s, and how many $\text{R}$ s ended up in the table.

Diagram

The first 4 columns of the 20×20 letter grid plus the last column (col 20). The P-Q-R cycle repeats every 3 rows (bracket at left); col 20 row 3 is highlighted because that position is also a P.

1
2
3
4
⋮
1
2
3
4
⋯

P
Q
R
P
⋯
Q
R
P
Q
⋯
R
P
Q
R
⋯
P
Q
R
P
⋯
⋮
⋮
⋮
⋮
⋱
20
⋯
⋯
⋯
⋯
⋮

Q

R

P

Q

⋮

repeats every 3 rows

20 columns

20 rows

= P cell

= Q or R

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.

Try it yourself first — the explanation is most useful after you’ve attempted it.

View mode:

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

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

#5 Look for a Pattern 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: $P, Q$.

Row $1: 6 + 1 = 7\text{ Ps},\; 6 + 1 = 7\text{ Qs},\; 6\text{ Rs}$

💡 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.

#5 Look for a Pattern 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: $Q, R$.

Row $2: 6\text{ Ps},\; 6 + 1 = 7\text{ Qs},\; 6 + 1 = 7\text{ Rs}$

💡 A shifted version of the same rule still produces a pattern — just the starting letter changes.

#5 Look for a Pattern 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 $R, P$.

Row $3: 6 + 1 = 7\text{ Ps},\; 6\text{ Qs},\; 6 + 1 = 7\text{ Rs}$

💡 Same rule, shifted again — Grade 4 pattern generation handles all three row types.

#9 Solve an Easier Related Problem 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.

$$P: 7+6+7 = 20,\;\; Q: 7+7+6 = 20,\;\; R: 6+7+7 = 20$$

💡 Solving the easier $3 \times 20$ block first and seeing $20$-$20$-$20$ is a Grade 3 'identify the arithmetic pattern' move.

#7 Identify Subproblems 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.

Rows $1\!-\!18: 6 \times 20 = 120\text{ Ps},\; 120\text{ Qs},\; 120\text{ Rs}$

💡 Multiplying $6 \times 20 = 120$ is a Grade 3 multiplication fact done three times.

#7 Identify Subproblems 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.

Rows $19, 20: (7+6)\text{ Ps} = 13,\;\; (7+7)\text{ Qs} = 14,\;\; (6+7)\text{ Rs} = 13$

💡 Splitting off the two leftover rows and adding their counts is a Grade 4 multi-step word-problem move.

#7 Identify Subproblems 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 $(C)$.

$$P: 120 + 13 = 133,\;\; Q: 120 + 14 = 134,\;\; R: 120 + 13 = 133 \;\Rightarrow\; \textbf{(C)}$$

💡 Recombining the easy pieces gives the final answer — the Grade 4 multi-step problem finish line.

[1] #5 4.OA.C.5 Count the letters in row $1$ by reading off the pattern $P, Q, R, P, Q, R, \dots

[2] #5 4.OA.C.5 Row $2$ is the same cycle but shifted by one, so it starts $Q, R, P, Q, R, P, \d

[3] #5 4.OA.C.5 Row $3$ is shifted by two, so it starts $R, P, Q, R, P, Q, \dots$. Six full $RPQ

[4] #9 3.OA.D.9 Add rows $1, 2, 3$ together to find the totals for one full $3 \times 20$ block.

[5] #7 3.OA.C.7 Row $4$ uses the same rule as row $1$ (shift by three is no shift), so the $3$-r

[6] #7 4.OA.A.3 Handle the leftover rows $19$ and $20$. Row $19$ behaves like row $1$ (since $19

[7] #7 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.7 Fluently multiply and divide within 100 (Computing $6 \times 20 = 120$ for the row counts across six full $3$-row blocks.)
3.OA.D.9 Identify 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.3 Solve 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.5 Generate 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!

Problem

Diagram

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 4)

More like this