Problem:
Consider polynomials P(x) of degree at most 3 , each of whose coefficients is an element of {0,1,2,3,4,5,6,7,8,9}. How many such polynomials satisfy P(−1)=−9 ?
Answer Choices:
A. 110
B. 143
C. 165
D. 220
E. 286
Solution:
Let P(x)=ax3+bx2+cx+d, where a,b,c, and d are integers between 0 and 9 , inclusive. The condition P(−1)=−9 is equivalent to −a+b−c+d=−9. Adding 18 to both sides gives (9−a)+b+(9−c)+d=9 where 0≤9−a,b,9−c,d≤9. By the stars and bars argument, there are (9+4−14−1)=(123)=220 nonnegative integer solutions to x1+x2+x3+x4=9. Each of these give rise to one of the desired polynomials.
OR
With the notation above, note that (a+c)−(b+d)=9 can occur in several ways: b+d=k,a+c=9+k where k=0,1,2,…,9. There are k+1 solutions to b+d=k and 10−k solutions to a+c=9+k under the restrictions on a,b,c, and d, yielding ∑k=09(k+1)(10−k)=220 solutions in all.
The problems on this page are the property of the MAA's American Mathematics Competitions