2025 AMC 10A Problem 19

The question has two parts: first find which row adds up to $12{,}288$, then find the third number of that row.

Part 1: which row.

1. Start by adding up the three rows in the picture:
- Row 1: $-1 + 3 + 1 = 3$
- Row 2: $-1 + 2 + 4 + 1 = 6$
- Row 3: $-1 + 1 + 6 + 5 + 1 = 12$

The sum doubles each time. Before trusting that, check why it happens.

2. Look at how row 3 is built from row 2, which is $-1, 2, 4, 1$. The middle numbers of row 3 are the sums of neighbours: $-1+2$, $2+4$, $4+1$. The inside numbers $2$ and $4$ each get used twice, and the end numbers $-1$ and $1$ get used once each. So these three sums add up to
$$2 \times 6 - (-1) - 1 = 12.$$
3. Then row 3 gets a new $-1$ at the start and a new $1$ at the end. Those two cancel, so the row sum is exactly $2 \times 6 = 12$.
4. The same thing happens in every row, because every row starts with $-1$ and ends with $1$. So the sums really do double: $3, 6, 12, 24, \ldots$ Row $n$ adds up to $3 \times 2^{\,n-1}$ (row 1 is $3 \times 1$, row 2 is $3 \times 2$, row 3 is $3 \times 4$).
5. Now $12{,}288 \div 3 = 4096 = 2^{12}$. So $n - 1 = 12$, which means the row is row 13, not row 12.

Part 2: the third number of row 13.

6. Writing out whole rows down to row 13 would be slow, since row 13 has $15$ numbers. Luckily the first three numbers of a row depend only on the first three numbers of the row above:
- the first number is always $-1$;
- the new second number is $-1 + (\text{old second number})$;
- the new third number is $(\text{old second number}) + (\text{old third number})$.
7. So the second number goes down by $1$ in every row, and the third number grows by adding the old second number. Make a table, starting from row 1 ($3$ and $1$):

1. Rows 2 and 3 match the picture ($2, 4$ and $1, 6$), so the table is set up correctly.
2. The third number of row 13 is $-29$.

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

When each row is built from the row above, you rarely need whole rows. Pick out just the things the question asks about (here the row sum and the first three numbers) and find the rule for how each one changes from one row to the next. The sum doubles because every number is used twice except the ends, and the ends are set up to cancel. The second number drops by $1$ each time, so the third number is just a running total.

Skip the table and add in one go. The third number starts at $1$ in row 1, and each step down adds the second number of the row above. To reach row 13 there are $12$ steps, which add the second numbers of rows 1 to 12:
$$3 + 2 + 1 + 0 + (-1) + (-2) + \cdots + (-8).$$
That is $12$ numbers going evenly from $3$ down to $-8$. Their average is $\frac{3 + (-8)}{2} = -2.5$, so their total is $12 \times (-2.5) = -30$. The third number of row 13 is $1 + (-30) = -29$.

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