The sum of several integers has a remainder 5 in division by 6. Find the remainder of the sum of the third powers of these numbers in division by 6

Answer 1

For any integer n the difference n³ - n = (n-1)n(n+1) is the product of three consecutive integers. One of these factors should be divisible by 3, and at least one factor should be divisible by 2.

So, n³ - n should be divisible by 6.

This means that (sum of third powers) - (sum of original numbers) is divisible by 6.

If the sum of numbers has the remainder 5 in division by 6, then the sum of their third powers has the same remainder 5 in division by 6.

Answer 2

This is problem best handled using modular arithmetic as explained in the above brief video.