2024 AMC 10A Problem 21

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 $\boxed{\textbf{(C)}\ 29}$.

"Every row and every column is arithmetic" is exactly the statement that the entry is linear in each index separately, which forces the bilinear form $\alpha + \beta i + \gamma j + \delta ij$ with four unknowns; four given entries determine it. Shifting the origin to the known zero entry removes one unknown before you start. This structure recurs in any grid whose rows and columns are all arithmetic.

• Assuming every row has the same common difference (a linear function Bi + Cj + A), which cannot fit the four given values; the row differences themselves form an arithmetic progression.
• Mixing up row and column indices when substituting the given positions (e.g. reading $(2,4)$ as column $2$, row $4$).