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

2011 AMC 12A Problem 20

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20.

Let

f(x) = ax^2 + bx + c,

where

a,

b,

and

c

are integers. Suppose that

f(1) = 0,

50 \lt f(7) \lt 60,

70 \lt f(8) \lt 80,

and

5000k \lt f(100) \lt 5000(k+1)

for some integer

k.

What is

k?

1

2

3

4

5

From

f(1) = 0,

c = -a - b,

f(7) = 6(8a + b)

and

f(8) = 7(9a + b)

The bounds force

8a + b = 9

and

9a + b = 11

Since

f(1) = a + b + c = 0,

we have

c = -a - b.

Then

f(7) = 48a + 6b = 6(8a + b),

f(8) = 63a + 7b = 7(9a + b).

From

50 \lt 6(8a + b) \lt 60

we get

8a + b = 9,

and from

70 \lt 7(9a + b) \lt 80

we get

9a + b = 11.

Subtracting,

a = 2,

then

b = -7

and

c = 5.

f(100) = 20000 - 700 + 5

= 19305,

which lies in

5000 \cdot 3 \lt 19305 \lt 5000 \cdot 4,

giving

k = 3.

Thus, the correct answer is C.