2025 AMC 10A Problem 19

Which row. Let row $n$ be $a_1, a_2, \ldots, a_m$ with $a_1 = -1$ and $a_m = 1$, and let $S_n$ be its sum. Row $n+1$ consists of $-1$, then the $m-1$ pairwise sums $a_1 + a_2, \ldots, a_{m-1} + a_m$, then $1$. In those pairwise sums every entry of row $n$ appears twice except the two ends, which appear once, so they total
$$2S_n - a_1 - a_m = 2S_n + 1 - 1 = 2S_n,$$
and the appended $-1$ and $1$ cancel. Hence
$$S_{n+1} = 2S_n.$$
With $S_1 = -1 + 3 + 1 = 3$ we get $S_n = 3 \cdot 2^{\,n-1}$; the printed rows check this ($3, 6, 12$). Since $12{,}288 = 3 \cdot 4096 = 3 \cdot 2^{12}$, we need $n - 1 = 12$, so it is row $13$.

The third entry. Write $b_n$ and $c_n$ for the second and third entries of row $n$. From the construction,
$$b_{n+1} = a_1 + b_n = b_n - 1, \qquad c_{n+1} = b_n + c_n.$$
Starting from $b_1 = 3$, the first recursion gives $b_n = 4 - n$. Then
$$c_n = c_1 + \sum_{j=1}^{n-1} b_j = 1 + \sum_{j=1}^{n-1}(4 - j) = 1 + 4(n-1) - \frac{(n-1)n}{2}.$$
Checking against the picture: $c_2 = 1 + 4 - 1 = 4$ and $c_3 = 1 + 8 - 3 = 6$, as printed.

For $n = 13$,
$$c_{13} = 1 + 4 \cdot 12 - \frac{12 \cdot 13}{2} = 1 + 48 - 78 = -29.$$

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

The two halves of the problem are the same trick applied twice: a rule that builds each row from the previous one gives a recursion, and a recursion whose step is constant or linear can be summed in closed form. The row sum obeys $S_{n+1} = 2S_n$ because the "sum adjacent pairs" operation always doubles a sum apart from the boundary, and here the boundary is rigged to contribute nothing. Only the first three entries of a row are ever needed, and each of them depends only on entries at least as far left, so a two-term recursion suffices; there is no need to write out whole rows. The same set-up governs any Pascal-like array with forced ends.

Tabulate only the second and third entries, one row per line, starting from $(3, 1)$:
$$(3,1),\ (2,4),\ (1,6),\ (0,7),\ (-1,7),\ (-2,6),\ (-3,4),\ (-4,1),\ (-5,-3),\ (-6,-8),\ (-7,-14),\ (-8,-21),\ (-9,-29).$$
The thirteenth pair gives $-29$. Note that the last five third entries are exactly the five answer choices in reverse, which is the examiners' way of punishing an off-by-one in the row number.

• Reading 12,288 = 3 * 2^12 as row 12 instead of row 13, which gives the third entry -21, choice (B).
• Assuming the row sums are $3, 6, 12, \ldots$ only by inspection of the three printed rows and never checking why; the doubling needs the fact that the two end entries are $-1$ and $1$.
• Recomputing whole rows down to row $13$, which is slow and invites an arithmetic slip in the growing middle entries.