What is the remainder when (1^3)+(2^3)+(3^3)+...+(1990^3) is divided by 7?
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Try to prove: If x is integer, then x^3(mod7)=0 or 1 or 6
Originally Posted by chris86 What is the remainder when (1^3)+(2^3)+(3^3)+...+(1990^3) is divided by 7? We need merely note that and that . From this we can note that It follows then that Note though that in we have that and so in particular Note though that in general And so in particular
Originally Posted by Drexel28 We need merely note that and that . From this we can note that It follows then that Note though that in we have that and so in particular Note though that in general And so in particular The above is too involved, imho. Using Zaratustra's hint, it's easy to show that , so since , we get that the sum equals Tonio
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