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

Notice that $\displaystyle P_{20}(x) = P_{19}(x - 20) = P_{18}((x - 20) - 19)$ $\displaystyle = P_{17}(((x - 20) - 19) - 18) \ldots$ $\displaystyle = P_0(x - (20 + 19 + 18 + \ldots + 2 + 1))$ . Using the formula for the sum of the first $n$ numbers, $1 + 2 \ldots + 20 = \frac{20(20+1)}{2} = 210$ . Thus, $\displaystyle P_{20}(x) = P_0(x - 210)$ .

Substitute $\displaystyle x - 210$ into the equation, so we get $\displaystyle (x - 210)^3 + 313(x - 210)^2 - 77(x - 210) - 8$ . The cubic will have a term of ${3\choose1}210^2x = 630 \cdot 210x$ . The square will have a term of $-313 \cdot {2\choose1}210x = -626 \cdot 210x$ . The linear part will have a term of $\displaystyle -77x$ . Adding up the coefficients, we get $630 \cdot 210 - 626 \cdot 210 - 77 = 763$ .

1993 AIME (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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1993 AIME Problems/Problem 5

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