Difference between revisions of "2000 AIME II Problems/Problem 14"

Revision as of 19:43, 11 November 2007

Problem

Every positive integer $k$ has a unique factorial base expansion $(f_1,f_2,f_3,\ldots,f_m)$ , meaning that $k=1!\cdot f_1+2!\cdot f_2+3!\cdot f_3+\cdots+m!\cdot f_m$ , where each $f_i$ is an integer, $0\le f_i\le i$ , and $0<f_m$ . Given that $(f_1,f_2,f_3,\ldots,f_j)$ is the factorial base expansion of $16!-32!+48!-64!+\cdots+1968!-1984!+2000!$ , find the value of $f_1-f_2+f_3-f_4+\cdots+(-1)^{j+1}f_j$ .

Solution

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 == Problem ==
-In triangle <math>ABC,</math> it is given that angles <math>B</math> and <math>C</math> are congruent. Points <math>P</math> and <math>Q</math> lie on <math>\overline{AC}</math> and <math>\overline{AB},</math> respectively, so that <math>AP = PQ = QB = BC.</math> Angle <math>ACB</math> is <math>r</math> times as large as angle <math>APQ,</math> where <math>r</math> is a positive real number. Find the greatest integer that does not exceed <math>1000r</math>.
+Every positive integer <math>k</math> has a unique factorial base expansion <math>(f_1,f_2,f_3,\ldots,f_m)</math>, meaning that <math>k=1!\cdot f_1+2!\cdot f_2+3!\cdot f_3+\cdots+m!\cdot f_m</math>, where each <math>f_i</math> is an integer, <math>0\le f_i\le i</math>, and <math>0<f_m</math>. Given that <math>(f_1,f_2,f_3,\ldots,f_j)</math> is the factorial base expansion of <math>16!-32!+48!-64!+\cdots+1968!-1984!+2000!</math>, find the value of <math>f_1-f_2+f_3-f_4+\cdots+(-1)^{j+1}f_j</math>.
 == Solution ==

2000 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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Difference between revisions of "2000 AIME II Problems/Problem 14"

Revision as of 19:43, 11 November 2007

Problem

Solution

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