2011 AMC 12A Problem 20

Attempt Problem 20 of the 2011 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 12A solutions, or check the answer key.

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20.

Let f(x)=ax2+bx+c,f(x) = ax^2 + bx + c, where a,a, b,b, and cc are integers. Suppose that f(1)=0,f(1) = 0, 50<f(7)<60,50 \lt f(7) \lt 60, 70<f(8)<80,70 \lt f(8) \lt 80, and 5000k<f(100)<5000(k+1)5000k \lt f(100) \lt 5000(k+1) for some integer k.k. What is k?k?

11

22

33

44

55

Answer: C
Concepts:quadraticsystem of equationsbounding to limit cases
Difficulty rating: 2030
Solution:

Since f(1)=a+b+c=0,f(1) = a + b + c = 0, we have c=ab.c = -a - b. Then f(7)=48a+6b=6(8a+b), f(7) = 48a + 6b = 6(8a + b), f(8)=63a+7b=7(9a+b). f(8) = 63a + 7b = 7(9a + b).

From 50<6(8a+b)<6050 \lt 6(8a + b) \lt 60 we get 8a+b=9,8a + b = 9, and from 70<7(9a+b)<8070 \lt 7(9a + b) \lt 80 we get 9a+b=11.9a + b = 11. Subtracting, a=2,a = 2, then b=7b = -7 and c=5.c = 5.

So f(100)=20000700+5f(100) = 20000 - 700 + 5 =19305,= 19305, which lies in 50003<19305<50004,5000 \cdot 3 \lt 19305 \lt 5000 \cdot 4, giving k=3.k = 3.

Thus, the correct answer is C.

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