2025 AMC 10B Problem 3

P1-10~3 minPrint

Sequences & Series Set up the equation/formula and compute; no special trick needed Compute small cases, spot the pattern, generalize

A Pascal-like triangle has $10$ as the top row and $10$ followed by $1$ as the second row. In each subsequent row the first number is $10$ , the last number is $1$ , and, as in the standard Pascal's Triangle, each other number in the row is the sum of the two numbers directly above it. The first four rows are shown below:

$\begin{array}{ccccccc} &&&10&&&\\ &&10&&1&& \\ &10&&11&&1& \\ 10&&21&&12&&1 \\ \end{array}$

What is the sum of the digits of the sum of the numbers in the $11$ th row?

Show solution without answering

Key insight

From the second row on, each row's total is exactly double the one before, so the totals go 11, 22, 44, … and the 11th row totals 5632.

Solution

Add up the rows you can see.

Row	Numbers	Total
1	$10$	$10$
2	$10, 1$	$11$
3	$10, 11, 1$	$22$
4	$10, 21, 12, 1$	$44$

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 $5632$ .

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

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

Why this works

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.1)
Intermediate Algebra, Chapter 17: More Sequences and Series (§17.1, §17.2)
Introduction to Number Theory, Chapter 12: Introduction to Modular Arithmetic (§12.6)