Difference between revisions of "1993 AIME Problems/Problem 5"
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== Problem == | == Problem == | ||
− | Let <math>P_0(x) = x^3 + 313x^2 - 77x - 8\,</math>. For | + | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Let <math>P_0(x) = x^3 + 313x^2 - 77x - 8\,</math>. For [[integer]]s <math>n \ge 1\,</math>, define <math>P_n(x) = P_{n - 1}(x - n)\,</math>. What is the [[coefficient]] of <math>x\,</math> in <math>P_{20}(x)\,</math>?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> |
− | == Solution == | + | == Solution 1 == |
− | {{ | + | Notice that |
+ | <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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+ | 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]]. | ||
+ | *<math>(x-210)^3</math> will have a linear term of <math>{3\choose1}210^2x = 630 \cdot 210x</math>. | ||
+ | *<math>313(x-210)^2</math> will have a linear term of <math>-313 \cdot {2\choose1}210x = -626 \cdot 210x</math>. | ||
+ | *<math>-77(x-210)</math> will have a linear term of <math>-77x</math>. | ||
+ | Adding up the coefficients, we get <math>630 \cdot 210 - 626 \cdot 210 - 77 = \boxed{763}</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | Notice the transformation of <math>P_{n-1}(x)\to P_n(x)</math> adds <math>n</math> to the roots. Thus, all these transformations will take the roots and add <math>1+2+\cdots+20=210</math> to them. (Indeed, this is very easy to check in general.) | ||
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+ | Let the roots be <math>r_1,r_2,r_3.</math> Then <math>P_{20}(x)=(x-r_1-210)(x-r_2-210)(x-r_3-210).</math> By Vieta's/expanding/common sense, you see the coefficient of <math>x</math> is <math>(r_1+210)(r_2+210)+(r_2+210)(r_3+210)+(r_3+210)(r_1+210).</math> Expanding yields <math>r_1r_2+r_2r_3+r_3r_1+210\cdot 2(r_1+r_2+r_3)+3\cdot 210^2.</math> Using Vieta's (again) and plugging stuff in yields <math>-77+210\cdot 2\cdot -313+3\cdot 210^2=\boxed{763}.</math> | ||
== See also == | == See also == | ||
{{AIME box|year=1993|num-b=4|num-a=6}} | {{AIME box|year=1993|num-b=4|num-a=6}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 13:25, 31 December 2018
Contents
Problem
Let . For integers , define . What is the coefficient of in ?
Solution 1
Notice that
Using the formula for the sum of the first numbers, . Therefore,
Substituting into the function definition, we get . We only need the coefficients of the linear terms, which we can find by the binomial theorem.
- will have a linear term of .
- will have a linear term of .
- will have a linear term of .
Adding up the coefficients, we get .
Solution 2
Notice the transformation of adds to the roots. Thus, all these transformations will take the roots and add to them. (Indeed, this is very easy to check in general.)
Let the roots be Then By Vieta's/expanding/common sense, you see the coefficient of is Expanding yields Using Vieta's (again) and plugging stuff in yields
See also
1993 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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