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Difference between revisions of "1993 AIME Problems/Problem 5"

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== Solution ==
 
== Solution ==
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Notice that <math>\displaystyle P_{20}(x) = P_{19}(x - 20) = P_{18}((x - 20) - 19)</math><math>\displaystyle = P_{17}(((x - 20) - 19) - 18) \ldots</math><math>\displaystyle = P_0(x - (20 + 19 + 18 + \ldots + 2 + 1))</math>. Using the formula for the sum of the first <math>n</math> numbers, <math>1 + 2 \ldots + 20 = \frac{20(20+1)}{2} = 210</math>. Thus, <math>\displaystyle P_{20}(x) = P_0(x - 210)</math>.
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Notice that  
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<cmath>\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*}</cmath>
  
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Substitute <math>\displaystyle x - 210</math> into the equation, so we get <math>\displaystyle (x - 210)^3 + 313(x - 210)^2 - 77(x - 210) - 8</math>. The cubic will have a term of <math>{3\choose1}210^2x = 630 \cdot 210x</math>. The square will have a term of <math>-313 \cdot {2\choose1}210x = -626 \cdot 210x</math>. The linear part will have a term of <math>\displaystyle -77x</math>. Adding up the coefficients, we get <math>630 \cdot 210 - 626 \cdot 210 - 77 = 763</math>.
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 +
Using the formula for the sum of the first <math>n</math> numbers, <math>1 + 2 + \cdots + 20 = \frac{20(20+1)}{2} = 210</math>. Therefore, <cmath>P_{20}(x) = P_0(x - 210).</cmath>
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Substituting <math>x - 210</math> into the function definition, we get <math>P_0(x-210) = (x - 210)^3 + 313(x - 210)^2 - 77(x - 210) - 8</math>. We only need the coefficients of the linear terms, which we can find by the [[binomial theorem]].
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*<math>(x-210)^3</math> will have a linear term of <math>{3\choose1}210^2x = 630 \cdot 210x</math>.
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*<math>313(x-210)^2</math> will have a linear term of <math>-313 \cdot {2\choose1}210x = -626 \cdot 210x</math>.  
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*<math>-77(x-210)</math> will have a linear term of <math>-77x</math>.  
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Adding up the coefficients, we get <math>630 \cdot 210 - 626 \cdot 210 - 77 = \boxed{763}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 22:36, 27 February 2008

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 \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 $630 \cdot 210 - 626 \cdot 210 - 77 = \boxed{763}$.

See also

1993 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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All AIME Problems and Solutions
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