2024 AMC 10A Problem 21

Four numbers are given, so we look for a formula for the grid with four unknowns. The corner $(5, 5)$ is $0$, so it is easiest to measure from there.

1. New labels. For the entry in row $i$, column $j$, let $u = 5 - i$ (how many rows above the bottom) and $v = 5 - j$ (how many columns left of the right edge). The $0$ is at $u = 0$, $v = 0$.
2. The right-hand column. Column $5$ is arithmetic and starts at $0$ at the bottom. So going up it reads $0, P, 2P, 3P, 4P$ for some step $P$: the entry at height $u$ is $Pu$.
3. The row steps. Each row is arithmetic, with its own step going left; call the step in row $u$ by the name $s_u$. Then $s_u$ is (column $4$ entry) minus (column $5$ entry) in that row. Both columns are arithmetic going up, so their difference also changes by the same amount each row. That means the row steps form an arithmetic sequence too: $s_u = Q + Ru$ for some numbers $Q$ and $R$.
4. The formula. Start at the column-5 entry $Pu$ and move $v$ steps left:
$$\text{entry} = Pu + v\,(Q + Ru) = Pu + Qv + Ruv.$$
5. Use the three other given numbers.
- Position $(2, 4)$: $u = 3$, $v = 1$. So $3P + Q + 3R = 48$.
- Position $(4, 3)$: $u = 1$, $v = 2$. So $P + 2Q + 2R = 16$.
- Position $(3, 1)$: $u = 2$, $v = 4$. So $2P + 4Q + 8R = 12$; halve it: $P + 2Q + 4R = 6$.
6. Solve. The last two equations differ only in the $R$ term: subtracting gives $2R = 6 - 16 = -10$, so $R = -5$.
Then the second equation becomes $P + 2Q = 26$, and the first becomes $3P + Q = 63$.
From $P = 26 - 2Q$: $3(26 - 2Q) + Q = 63$, so $78 - 5Q = 63$, $Q = 3$, and $P = 20$.
7. Check the three given numbers: $60 + 3 - 15 = 48$, $20 + 6 - 10 = 16$, $40 + 12 - 40 = 12$. All correct.
8. Position $(1, 2)$: $u = 4$, $v = 3$:
$$20 \cdot 4 + 3 \cdot 3 + (-5) \cdot 4 \cdot 3 = 80 + 9 - 60 = 29.$$

The answer is $\boxed{\textbf{(C)}\ 29}$.

When every row and every column is arithmetic, the whole grid is fixed by just four numbers: the steps can change from row to row, but only by a steady amount. Writing the entry as a formula with four unknowns and plugging in the four given entries does the rest. Putting the starting point at the given $0$ removes one unknown before you begin.

Let $x(i,j)$ be the entry in row $i$, column $j$. Row $i$ is arithmetic, so $x(i,j) = x(i,5) + (j-5)\,d_i$ where $d_i$ is that row's common difference. Column $5$ is arithmetic, so $x(i,5)$ is a linear function of $i$. Also $x(i,4) - x(i,5) = -d_i$ is the difference of two columns, each arithmetic in $i$, so $d_i$ is linear in $i$ as well. Therefore every entry has the form
$$x(i,j) = \alpha + \beta i + \gamma j + \delta\, ij.$$

Measure from the corner: set $u = 5 - i$, $v = 5 - j$. Since $x(5,5) = 0$ the constant term vanishes and $x = Pu + Qv + Ruv$. The three other data points give

• $(2,4)$, i.e. $u = 3, v = 1$: $3P + Q + 3R = 48$,
• $(4,3)$, i.e. $u = 1, v = 2$: $P + 2Q + 2R = 16$,
• $(3,1)$, i.e. $u = 2, v = 4$: $2P + 4Q + 8R = 12$, or $P + 2Q + 4R = 6$.

Subtracting the second from the third: $2R = -10$, so $R = -5$. Then $P + 2Q = 26$ and $3P + Q = 63$; from the first $P = 26 - 2Q$, so $78 - 5Q = 63$, giving $Q = 3$ and $P = 20$.

Position $(1,2)$ has $u = 4$, $v = 3$:
$$x = 20\cdot 4 + 3\cdot 3 + (-5)(4)(3) = 80 + 9 - 60 = 29.$$

The answer is $\boxed{\textbf{(C)}\ 29}$.