Give the last three digits of the sum of cubes of the first 2016 natural numbers.

The answer is 496.

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Sum of the cubes of the first n numbers: (n(n+1)/2)^2 Plug in 2016 we obtain, 2033136^2 Under modulo 1000 (2033136^2)mod1000=== 136^2mod1000= 496mod1000