AMC 8 · 2007 · #9

Easy mode Grade 4

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Problem

Imagine a $4 \times 4$ grid. Some squares are filled in, and the rest are blank.

The rule: in every row, the digits $1, 2, 3, 4$ must each appear exactly once. The same rule holds for every column.

Here is the grid:

$\begin{tabular}{|c|c|c|c|}\hline 1 & & 2 &\\ \hline 2 & 3 & &\\ \hline & &&4\\ \hline & &&\\ \hline\end{tabular}$

What number goes in the lower right square (the very bottom-right corner)?

Pick an answer.

(A)

$1$

(B)

$2$

(C)

$3$

(D)

$4$

(E)

cannot be determined

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.

View mode:

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

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

#1 Draw a Diagram 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.

$$\begin{array}{|c|c|c|c|} \hline 1 & & 2 & ? \\ \hline 2 & 3 & & ? \\ \hline & & & 4 \\ \hline & & & ? \\ \hline \end{array}$$

💡 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.

#3 Eliminate Possibilities 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.

$\{(1,4),\,(2,4),\,(4,4)\} = \{1,\,2,\,3\}$ (in some order)

💡 Removing the digit already placed in the column shrinks the candidate set for every empty cell to $\{1, 2, 3\}$.

#3 Eliminate Possibilities 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, $(4,4)$.

$$(1,4) \ne 2,\;(2,4) \ne 2,\;(3,4) = 4 \;\Rightarrow\; (4,4) = 2$$

💡 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 $4$.

#3 Eliminate Possibilities 4.OA.C.5 Step 4

Conclude that $(4,4) = 2$, which is answer choice (B).

$$(4,4) = 2 \;\Rightarrow\; \textbf{(B)}$$

💡 Only one digit survived elimination, so the answer is determined — choice (E) "cannot be determined" is ruled out.

[1] #1 3.OA.D.9 Draw the grid and mark the four cells of column $4$. The cell we want is $(4,4)$

[2] #3 4.OA.C.5 Cell $(3,4)$ is already $4$, so column $4$ has its $4$. That means $(1,4)$, $(2,

[3] #3 4.OA.C.5 Find where the digit $2$ can sit in column $4$. Check each empty cell against it

[4] #3 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.9 Identify 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.5 Generate 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.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 4)

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