1993 AIME Problem 5

Attempt Problem 5 of the 1993 AIME below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1993 AIME solutions, or check the answer key.

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5.

Let P0(x)=x3+313x2−77x−8.P_0(x)=x^3+313x^2-77x-8. For integers n≥1,n\geq1, define Pn(x)=Pn−1(x−n).P_n(x)=P_{n-1}(x-n). What is the coefficient of xx in P20(x)?P_{20}(x)?
Answer: 763
Concepts:algebraic manipulationbinomial theorempolynomial
Small hint:
Collapse the repeated shifts to write P20P_{20} directly in terms of P0P_0
Big hint:
Only collect the linear terms after substituting x−210x-210
Solution:
The accumulated shift is 1+2+⋯+20=210,1+2+\cdots+20=210, so P20(x)=P0(x−210).P_{20}(x)=P_0(x-210). The coefficient of xx is therefore 3(210)2−2(313)(210)−77=132300−131460−77=763.\begin{aligned}&3(210)^2-2(313)(210)-77\\&\quad=132300-131460-77\\&\quad=763.\end{aligned}
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