2025 AMC 10B Problem 3

Question

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.

What is the sum of the digits of the sum of the numbers in the

11

th row?

11

13

14

16

17

Answer: D

Concepts:Pascal’s Triangle recursion geometric sequence

Difficulty rating: 1270

Small hint:

Track only the row sums. Each interior entry is counted by both of the two entries below it

Big hint:

Show

S_n = 2 S_{n-1}

for

n \ge 3

with

S_2 = 11,

then compute

S_{11}

Solution:

Let

S_n

be the sum of row

n.

Each entry in row

n-1

feeds the two entries just below it, and the fixed border numbers

10

and

1

exactly make up for the terms lost at the edges. So

S_n = 2 S_{n-1}

for

n \ge 3.

With

S_2 = 11,

this gives

S_n = 11 \cdot 2^{n-2},

so

S_{11} = 11 \cdot 2^9 = 5632.

Its digits sum to

5 + 6 + 3 + 2 = 16.

Thus, D is the correct answer.