AMC 8 · 2007 · #9
Grade 4 logicProblem
To complete the grid below, each of the digits 1 through 4 must occur once
in each row and once in each column. What number will occupy the lower
right-hand square?
\begin{tabular}{|c|c|c|c|}\hline 1 & & 2 &\\ \hline 2 & 3 & &\\ \hline & &&4\\ \hline & &&\\ \hline\end{tabular}
Pick an answer.
AMC 8 2007 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 $4 \times 4$ grid is being filled so that each of the digits $1$, $2$, $3$, $4$ appears exactly once in every row and once in every column. Some cells are already filled. Find the digit that goes in the lower right-hand cell.
Givens: The grid is $4 \times 4$; Row $1$ has $1$ in column $1$ and $2$ in column $3$; Row $2$ has $2$ in column $1$ and $3$ in column $2$; Row $3$ has $4$ in column $4$; Every row uses each digit $1$, $2$, $3$, $4$ exactly once; Every column uses each digit $1$, $2$, $3$, $4$ exactly once; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) cannot be determined
Unknowns: The digit in cell $(4,4)$ — the lower right-hand square
Understand
Restated: A $4 \times 4$ grid is being filled so that each of the digits $1$, $2$, $3$, $4$ appears exactly once in every row and once in every column. Some cells are already filled. Find the digit that goes in the lower right-hand cell.
Givens: The grid is $4 \times 4$; Row $1$ has $1$ in column $1$ and $2$ in column $3$; Row $2$ has $2$ in column $1$ and $3$ in column $2$; Row $3$ has $4$ in column $4$; Every row uses each digit $1$, $2$, $3$, $4$ exactly once; Every column uses each digit $1$, $2$, $3$, $4$ exactly once; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) cannot be determined
Plan
Primary tool: #3 Eliminate Possibilities
Secondary: #1 Draw a Diagram
Cell $(4,4)$ lives in column $4$. By the rule, column $4$ must contain all four digits $1$, $2$, $3$, $4$. Three of its cells are either filled or share a row with one of the digits. Tool #3 (Eliminate Possibilities) lets us cross off three candidates for $(4,4)$ until only one digit is left. Tool #1 (Draw a Diagram) keeps a clean picture of the grid so we can see which row and column conflicts apply to each cell. No algebra is needed — only careful row/column bookkeeping.
Execute — Answer: B
3.OA.D.9 Step 1 - Draw the grid and mark the four cells of column $4$.
- The cell we want is $(4,4)$.
- We will work out what each of the other three column-$4$ cells must contain (or what they cannot contain) by checking their rows.
💡 Putting the grid on the page makes the row/column rule visible — each row needs the four digits exactly once, and so does each column. This is the Grade $3$ pattern-recognition move.
4.OA.C.5 Step 2 - Cell $(3,4)$ is already $4$, so column $4$ has its $4$.
- That means $(1,4)$, $(2,4)$, and $(4,4)$ must hold the remaining digits $1$, $2$, $3$ in some order.
💡 Removing the digit already placed in the column shrinks the candidate set for every empty cell to $\{1, 2, 3\}$.
4.OA.C.5 Step 3 - Find where the digit $2$ can sit in column $4$.
- Check each empty cell against its row.
- Row $1$ already has a $2$ in cell $(1,3)$, so $(1,4) \ne 2$.
- Row $2$ already has a $2$ in cell $(2,1)$, so $(2,4) \ne 2$.
- The only cell of column $4$ that can still take a $2$ is the bottom one, $(4,4)$.
💡 Process of elimination on a single column: cross off every cell where the digit cannot go, and the one cell left wins. The digit $2$ has only one legal home in column $4$.
4.OA.C.5 Step 4 Conclude that $(4,4) = 2$, which is answer choice (B).
💡 Only one digit survived elimination, so the answer is determined — choice (E) "cannot be determined" is ruled out.
3.OA.D.9 Draw the grid and mark the four cells of column $4$. The cell we want is $(4,4)$ 4.OA.C.5 Cell $(3,4)$ is already $4$, so column $4$ has its $4$. That means $(1,4)$, $(2, 4.OA.C.5 Find where the digit $2$ can sit in column $4$. Check each empty cell against it 4.OA.C.5 Conclude that $(4,4) = 2$, which is answer choice (B). Review
Reasonableness: Finish filling the grid to confirm the rule still holds. Row $1$ needs $3$ and $4$ in cells $(1,2)$ and $(1,4)$. Column $2$ already has $3$ at $(2,2)$, so $(1,2) \ne 3$, forcing $(1,2) = 4$ and $(1,4) = 3$. Then column $4$ has $3$ at $(1,4)$, $4$ at $(3,4)$, and $2$ at $(4,4)$, so $(2,4)$ must be $1$. Every row and column ends up using $1$, $2$, $3$, $4$ exactly once, and the lower right cell is indeed $2$.
Alternative: Tool #1 (Draw a Diagram) carried further: fill the whole grid by deduction. Row $1$ needs $3$ and $4$, but column $2$ already has $3$, so row $1$ reads $1, 4, 2, 3$. Row $2$ needs $1$ and $4$, but column $3$ will need a $1$ from row $2$, so row $2$ reads $2, 3, 4, 1$. Row $3$ needs $1$, $2$, $3$ in the first three columns; column $1$ already has $1$ and $2$, so $(3,1) = 3$, and the rest of row $3$ becomes $3, 1, 2, 4$. Row $4$ is whatever is left: $4, 2, 3, ?$. The only digit not yet used in column $4$ is $2$, confirming $(4,4) = 2$.
CCSS standards used (min grade 4)
3.OA.D.9Identify arithmetic patterns and explain them using properties of operations (Recognizing the Latin-square pattern that each row and each column must contain the four digits $1$, $2$, $3$, $4$ exactly once.)4.OA.C.5Generate a number or shape pattern that follows a given rule and identify apparent features of the pattern (Applying the row/column rule to eliminate impossible digits at each cell of column $4$, leaving exactly one valid digit for the lower right square.)
⭐ Focus on the column that holds the unknown cell. Cross off every digit blocked by its row or by another column entry, and the one digit left is your answer.
⭐ Focus on the column that holds the unknown cell. Cross off every digit blocked by its row or by another column entry, and the one digit left is your answer.
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