<prompt> Acha a_1, a_2, lddots, a_2005 ziwe nambari halisi ili anzearrayccca_1cdot 1 a_2 cdot 2 a_3 cdot 3 cdots a_2005 cdot 2005 0 a_1cdot 12 a_2cdot 22 a_3 cdot 32 cdots a_2005 cdot 20052 0 a_1 cdot 13 a_2 cdot 23 a_3 cdot 33 cdots a_2005 cdot 20053 0 vdotsvdots vdotsvdotsvdots a_1cdot 12004 a_2cdot 22004 a_3cdot 32004 cdots a_2005 cdot 20052004 0endarrayna anzaarrayccca_1 cdot 12005 a_2cdot 22005 a_3cdot 32005 cdots a_2005 cdot 20052005 1.endarrayThamani ya a_1 ni nini?  <response>: Kumbuka kuwa mlinganyo wa nth una uwezo wa nth, haswa 1n, 2n, dots, 2005n. Hii inatufanya tufikirie kutathmini baadhi ya polynomial px saa x 1, 2, dots, 2015. Swali ni ni polynomial ipi. Kwa hiyo, basipx c_2005 x2005 c_2004 x2004 dots c_1 x.Tukizidisha mlinganyo wa nth kwa c_n, basi tunapataanzasafuccca_1 cdot c_1 cdot 1 a_2 cdot c_1 cdot 2 a_3 cdot c_1 cdot 3 dotsb a_2005 cdot c_1 cdot 2005 0, a_1 cdot c_2 cdot 12 a_2 cdot c_2 cdot 22 a_3 cdot c_2 cdot 32 dotsb a_2005 cdot c_2 cdot 20052 0, a_1 cdot c_3 cdot 13 a_2 cdot c_2 cdot 23 a_3 cdot c_3 cdot 33 dotsb a_2005 cdot c_3 cdot 20053 0, vidoti, a_1 cdot c_2004 cdot 12004 a_2 cdot c_2 cdot 22004 a_3 cdot c_2004 cdot 32004 dotsb a_2005 cdot c_2004 cdot 20052004 0, a_1 cdot c_2005 cdot 12005 a_2 cdot c_2 cdot 22005 a_3 cdot c_2005 cdot 32005 dotsb a_2005 cdot c_2005 cdot 20052005 c_2005.mwishosafuKumbuka kwamba masharti katika safu wima kth yanaongeza hadi pk. Hivyo,a_1 p1 a_2 p2 a_3 p3 dots a_2005 p2005 c_2005.Kumbuka kwamba hii inashikilia kwa vifungu vyovyote c_1, c_2,dots,c_2005 tunayochagua. Kwa kuwa tunataka a_1, tunachagua viambajengo c_i ili maneno yote katika mlingano ulio hapo juu yatoweke, isipokuwa a_1 p1. Tunaweza kufanikisha hili kwa kuwekapx xx - 2x - 3 dotsm x - 2005.Kisha p1 2004! na pk 0 kwa k 2,, 3, dots, 2005, hivyo2004! cdot a_1 1.Kwa hivyo, a_1 frac12004!. 