Question

Show that the cube of a positive integer is of the form 6q + r, where q is an integer and r = 0, 1, 2, 3, 4, 5.

Answer 1

Given: Let n be any positive integer. By the division algorithm write n = 6q + r, where q is an integer and 0 ≤ r ≤ 5.

Expand and rearrange: n³ = (6q + r)³

= 216q³ + 108q²r + 18qr² + r³

= 6(36q³ + 18q²r + 3qr²) + r³

Thus n³ = 6M + r³ for the integer M = 36q³ + 18q²r + 3qr².

Now note that for each r = 0, 1, 2, 3, 4, 5 we have r³ ≡ r (mod 6) (equivalently r³ – r = r(r – 1)(r + 1) is divisible by 6), so r³ = 6k + r for some integer k.

Substituting gives n³ = 6(M + k) + r, i.e. n³ = 6Q + r for some integer Q.

Therefore, the cube of any positive integer is of the form 6q' + r with integer q' and r ∈ {0, 1, 2, 3, 4, 5}.