Problem:
Let f(x)=ax2+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 ?
Answer Choices:
A. 1
B. 2
C. 3
D. 4
E. 5
Solution:
Note that f(1)=a+b+c=0, so f(7)=49a+7b+c=48a+6b= 6(8a+b). Thus f(7) is an integer multiple of 6 strictly between 50 and 60 , so f(7)=54 and 8a+b=9. Similarly, f(8)=64a+8b+c=63a+7b=7(9a+b). Thus f(8) is an integer multiple of 7 strictly between 70 and 80 , so f(8)=77 and 9a+b=11. It follows that a=2,b=−7, and c=5. Therefore f(100)=2⋅1002−7⋅100+5=19,305, and thus k=3​.
The problems on this page are the property of the MAA's American Mathematics Competitions