2025 AMC 10B Problem 3

Let $S_n$ be the sum of row $n$. From the picture, $S_1 = 10$, $S_2 = 11$, $S_3 = 10 + 11 + 1 = 22$, and $S_4 = 10 + 21 + 12 + 1 = 44$.

The doubling is exact, and it is worth seeing why. Suppose row $n$ (with $n \ge 2$) is $a_1, a_2, \dots, a_k$, where $a_1 = 10$ and $a_k = 1$. Row $n+1$ is
$$10,\ a_1 + a_2,\ a_2 + a_3,\ \dots,\ a_{k-1} + a_k,\ 1.$$
In the middle entries every $a_i$ appears twice except $a_1$ and $a_k$, which appear once each, so those middle entries total $2S_n - a_1 - a_k = 2S_n - 11$. Adding the two ends back,
$$S_{n+1} = 10 + (2S_n - 11) + 1 = 2S_n.$$

So from row $2$ onwards the sums double: $S_n = 11 \cdot 2^{\,n-2}$. For $n = 11$,
$$S_{11} = 11 \cdot 2^{9} = 11 \cdot 512 = 5632,$$
and the digits sum to $5 + 6 + 3 + 2 = 16$.

The answer is $\boxed{\textbf{(D)}\ 16}$.

The doubling rule for Pascal's triangle survives unchanged here because it depends only on each interior entry being the sum of the two above it and the two end entries being copied down; the actual values of the ends are irrelevant to the ratio. What the altered corner does change is the starting value, which is why the formula must be anchored at row $2$ rather than row $1$. Writing out four row sums and checking $10, 11, 22, 44$ against the claimed formula is the fastest safeguard against an index slip.

Track the rows directly: $10, 11, 22, 44, 88, 176, 352, 704, 1408, 2816, 5632$. Under contest conditions this is about ten doublings and takes well under a minute, with no indexing to get wrong.

• Indexing the doubling from the first row instead of the second and computing 11 * 2^10 = 11264, whose digit sum 14 is choice (C).
• Doubling from $S_1 = 10$ to get $10 \cdot 2^{10} = 10240$; the first step is $10 \to 11$, not $10 \to 20$, because row $2$ has no interior entries.
• Answering $5632$ or stopping at $S_{11}$ without taking the digit sum.