Using the sum of cubes formula or modular arithmetic: 16^3 + 17^3 + 18^3 + 19^3. 16+19=35 and 17+18=35. Since a^3+b^3 is divisible by a+b, the sum is divisible by 35. Also, the sum of four consecutive cubes is divisible by 4. Thus it is divisible by 70, leaving remainder 0.