Modulus Operations Homework: Simplifying 1^3+2^3+3^3+...+99^3+100^3(mod4)

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To simplify the expression 1^3 + 2^3 + 3^3 + ... + 100^3 (mod 4), focus on the values of b from 1 to 4, as higher numbers will repeat their residues when cubed. The cubes of these residues are calculated as follows: 1^3 mod 4 = 1, 2^3 mod 4 = 0, 3^3 mod 4 = 3, and 4^3 mod 4 = 0. Each complete set of four numbers contributes 1 + 0 + 3 + 0 = 4, which is congruent to 0 mod 4. Therefore, since there are 25 complete sets of 4 in 100, the entire sum simplifies to 0 mod 4.
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How do you simplify : 1^3+2^3+3^3+4^3+...+99^3+100^3(mod4)

Please try to explain the solution as detailed as possible or atleast so I can understand it. :smile:
 
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Use the fact that (4a+b)^3 mod 4 = b^3 mod 4. All you have to worry about are b=1,2,3,4. All of the other terms in the sum are duplicates.
 
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