2011 AMC 10A Problem 17

Below is the professionally curated solution for Problem 17 of the 2011 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 10A solutions, or check the answer key.

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Concepts:recursionpattern recognition

Difficulty rating: 1370

17.

In the eight term sequence A,A, B,B, C,C, D,D, E,E, F,F, G,G, H,H, the value of CC is 55 and the sum of any three consecutive terms is 30.30. What is A+H?A+H?

1717

1818

2525

2626

4343

Solution:

From the condition about the sequence, we get that A+B+C=30 A + B + C = 30 B=25A. B = 25 - A. Similarly, we get B+C+D=30 B + C + D = 30 D=A. D = A.

Propagating these values through the sequence and repeating the condition for every consecutive triple, we get that E=25A,F=5,G=A, E = 25 - A, F = 5, G = A, and finally, H=25A.H = 25 - A.

The desired sum is then A+25A=25. A + 25 - A = 25.

Thus, C is the correct answer.

Problem 17 in Other Years