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2018 AMC 12B Problem 22

Problem 22 of 25HarderAlgebraCounting & Probability

Consider polynomials P(x)P(x) of degree at most 3,3, each of whose coefficients is an element of {0,1,2,3,4,5,6,7,8,9}.\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}. How many such polynomials satisfy P(1)=9?P(-1)=-9?

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Solution

Write P(x)=ax3+bx2+cx+dP(x)=ax^3+bx^2+cx+d with each of a,b,c,da,b,c,d in {0,,9}.\{0,\ldots,9\}. The condition is a+bc+d=9.-a+b-c+d=-9. Let a=9aa'=9-a and c=9c,c'=9-c, both in [0,9].[0,9]. Then a+b+c+d=9.a'+b+c'+d=9. By stars and bars the number of nonnegative solutions is (9+33)=(123)=220,\binom{9+3}{3}=\binom{12}{3}=220, and each automatically satisfies the upper bounds since the sum is 9.9. Thus, the correct answer is D.

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Concepts: stars and bars · substitution

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.