Algebra Level 3

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

The answer is 496.

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2 solutions

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

Feb 1, 2016

The formula for sum of cubes is 1/2[(n)(n+1)]^2. This will give you [(2016/2)*2017]^2, or (1008 x 2017)^2. Quickly squaring 1008 gives you 1016064, and squaring 2017 gives you 4034289. Since you are only looking for the last three digits, multiplying 64 by 289 will yield 18496, thus the last three digits are 496. I apologize for the formatting as this is my first solution post, and I'm still trying to learn LaTex.

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