1993 AIME Problems/Problem 5

Problem

Let $P_0(x) = x^3 + 313x^2 - 77x - 8\,$. For integers $n \ge 1\,$, define $P_n(x) = P_{n - 1}(x - n)\,$. What is the coefficient of $x\,$ in $P_{20}(x)\,$?

Solution 1

Notice that \begin{align*}P_{20}(x) &= P_{19}(x - 20)\\ &= P_{18}((x - 20) - 19)\\ &= P_{17}(((x - 20) - 19) - 18)\\ &= \cdots\\ &= P_0(x - (20 + 19 + 18 + \ldots + 2 + 1)).\end{align*}


Using the formula for the sum of the first $n$ numbers, $1 + 2 + \cdots + 20 = \frac{20(20+1)}{2} = 210$. Therefore, \[P_{20}(x) = P_0(x - 210).\]

Substituting $x - 210$ into the function definition, we get $P_0(x-210) = (x - 210)^3 + 313(x - 210)^2 - 77(x - 210) - 8$. We only need the coefficients of the linear terms, which we can find by the binomial theorem.

  • $(x-210)^3$ will have a linear term of ${3\choose1}210^2x = 630 \cdot 210x$.
  • $313(x-210)^2$ will have a linear term of $-313 \cdot {2\choose1}210x = -626 \cdot 210x$.
  • $-77(x-210)$ will have a linear term of $-77x$.

Adding up the coefficients, we get $630 \cdot 210 - 626 \cdot 210 - 77 = \boxed{763}$.

Solution 2

Notice the transformation of $P_{n-1}(x)\to P_n(x)$ adds $n$ to the roots. Thus, all these transformations will take the roots and add $1+2+\cdots+20=210$ to them. (Indeed, this is very easy to check in general.)

Let the roots be $r_1,r_2,r_3.$ Then $P_{20}(x)=(x-r_1-210)(x-r_2-210)(x-r_3-210).$ By Vieta's/expanding/common sense, you see the coefficient of $x$ is $(r_1+210)(r_2+210)+(r_2+210)(r_3+210)+(r_3+210)(r_1+210).$ Expanding yields $r_1r_2+r_2r_3+r_3r_1+210\cdot 2(r_1+r_2+r_3)+3\cdot 210^2.$ Using Vieta's (again) and plugging stuff in yields $-77+210\cdot 2\cdot -313+3\cdot 210^2=\boxed{763}.$

See also

1993 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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All AIME Problems and Solutions

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