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Algebra Difficulty 4.9 AIME Prove it Saudi Arabia

Find positive integers a1<a2<<a2010a_{1} < a_{2} < \ldots < a_{2010} such that
a1(1!)2010+a2(2!)2010++a2010(2010!)2010=(2011!)2010. a_{1}(1!)^{2010} + a_{2}(2!)^{2010} + \ldots + a_{2010}(2010!)^{2010} = (2011!)^{2010}.

Solution

Consider
a1=22010, a2=320101, a3=420101, , a2010=201120101a_{1} = 2^{2010},\ a_{2} = 3^{2010} - 1,\ a_{3} = 4^{2010} - 1,\ \ldots,\ a_{2010} = 2011^{2010} - 1
and get
a1(1!)2010+a2(2!)2010++a2010(2010!)2010=(2!)2010+(3!)2010(2!)2010+(4!)2010(3!)2010+++(2011!)2010(2010!)2010=(2011!)2010 \begin{gathered} a_{1}(1!)^{2010} + a_{2}(2!)^{2010} + \ldots + a_{2010}(2010!)^{2010} \\ = (2!)^{2010} + (3!)^{2010} - (2!)^{2010} + (4!)^{2010} - (3!)^{2010} + \ldots + \\ + (2011!)^{2010} - (2010!)^{2010} = (2011!)^{2010} \end{gathered}
Moreover, it is clear that a1<a2<a3<<a2010a_{1} < a_{2} < a_{3} < \ldots < a_{2010}.

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