Difference between revisions of "1993 AIME Problems/Problem 5"
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<!-- 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> | <!-- 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 | 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> | <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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*<math>-77(x-210)</math> will have a linear term of <math>-77x</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>. | 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.) | ||
| + | |||
| + | 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 == | ||
Revision as of 12:25, 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, ![\[P_{20}(x) = P_0(x - 210).\]](http://latex.artofproblemsolving.com/6/3/f/63f8ce4cbf48964dd61df68155e7c6b185964047.png)
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 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
