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2011 AMC 12A Problem 20

Problem 20 of 25HarderAlgebraProblem-Solving Techniques

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?

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Solution

Since f(1)=a+b+c=0,f(1) = a + b + c = 0, we have c=−a−b.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)=20000−700+5f(100) = 20000 - 700 + 5 =19305,= 19305, which lies in 5000⋅3<19305<5000⋅4,5000 \cdot 3 \lt 19305 \lt 5000 \cdot 4, giving k=3.k = 3. Thus, the correct answer is C.
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Tagged: quadratic · system of equations · bounding to limit cases

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