Restul impartirii unei sume de cuburi

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katos
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Restul impartirii unei sume de cuburi

Post by katos »

Fie A=1³ +2³ +3³ +...+2004³ . Care este restul impartirii lui A la 2009?
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Laurian Filip
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Post by Laurian Filip »

A=\( 1^3+2^3+3^3+4^3+(5^3+2004^3)+(6^3+2003^3)+...+(1004^3+1005^3) \)

Fiecare paranteza este divizibila cu 2009, deci restul e chiar \( 1^3+2^3+3^3+4^3. \)
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