2025 AMC 10B Problem 3

Below is the professionally curated solution for Problem 3 of the 2025 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AMC 10B solutions, or check the answer key.

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Concepts:Pascal’s Trianglerecursiongeometric sequence

Difficulty rating: 1270

3.

A Pascal-like triangle has 1010 as the top row and 1010 followed by 11 as the second row. In each subsequent row the first number is 10,10, the last number is 1,1, and, as in the standard Pascal 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 1111th row?

1111

1313

1414

1616

1717

Solution:

Let SnS_n be the sum of row n.n. Each entry in row n1n-1 feeds the two entries just below it, and the fixed border numbers 1010 and 11 exactly make up for the terms lost at the edges. So Sn=2Sn1S_n = 2 S_{n-1} for n3.n \ge 3. With S2=11,S_2 = 11, this gives Sn=112n2,S_n = 11 \cdot 2^{n-2}, so S11=1129=5632.S_{11} = 11 \cdot 2^9 = 5632. Its digits sum to 5+6+3+2=16.5 + 6 + 3 + 2 = 16. Thus, D is the correct answer.

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