2025 AMC 10B Problem 3

1. Add up the rows you can see.

From row $2$ on, each total is double the one before.

2. Check that the doubling really keeps going. Take row $4$: $10, 21, 12, 1$. Row $5$ is
$$10,\quad 10+21,\quad 21+12,\quad 12+1,\quad 1.$$
In the middle entries, $21$ and $12$ are each used twice, while the end numbers $10$ and $1$ are each used once. The new row also starts with a fresh $10$ and ends with a fresh $1$. So the end numbers are also counted twice in total, and every number of row $4$ is counted exactly twice. Row $5$ totals $2 \times 44 = 88$.

The same thing happens for every row after row $2$, because each row starts with $10$ and ends with $1$. (Row $1$ is different: it has no $1$ at the end, which is why the doubling starts at row $2$.)

3. Keep doubling up to row 11.
$$11,\ 22,\ 44,\ 88,\ 176,\ 352,\ 704,\ 1408,\ 2816,\ 5632$$
These are rows $2$ through $11$, so the 11th row totals $5632$.

1. The digit sum is $5 + 6 + 3 + 2 = 16$.

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

In any Pascal-style triangle, each number is used by the two numbers just below it, so the next row counts everything twice. That only works when the new row's end numbers match the old row's ends. Here that is true from row $2$ on, but not from row $1$ to row $2$. Writing out a few row totals by hand is the quickest way to see where the pattern starts.

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 $\boxed{\textbf{(D)}\ 16}$.