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

Introduction

In this problem, we study the possible forms of the cube of a positive integer. We will show that the cube of any positive integer can always be written in the form 6q + r, where q is an integer and r takes one of the values 0, 1, 2, 3, 4, or 5.

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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, or 5.

Solution

Let n be any positive integer. Every positive integer can be written in one of the following six forms:

6q, 6q + 1, 6q + 2, 6q + 3, 6q + 4, or 6q + 5, where q is an integer.

We now find the cube of n in each case.

If n = 6q, then n³ = (6q)³ = 216q³, which is divisible by 6. Hence, the cube is of the form 6q.

If n = 6q + 1, then n³ = (6q + 1)³ = 216q³ + 108q² + 18q + 1, which is of the form 6q + 1.

If n = 6q + 2, then n³ = (6q + 2)³ = 216q³ + 216q² + 72q + 8, which is of the form 6q + 2.

If n = 6q + 3, then n³ = (6q + 3)³ = 216q³ + 324q² + 162q + 27, which is of the form 6q + 3.

If n = 6q + 4, then n³ = (6q + 4)³ = 216q³ + 432q² + 288q + 64, which is of the form 6q + 4.

If n = 6q + 5, then n³ = (6q + 5)³ = 216q³ + 540q² + 450q + 125, which is of the form 6q + 5.

Thus, in all cases, the cube of a positive integer can be written in the form 6q + r, where r is one of the numbers 0, 1, 2, 3, 4, or 5.

Conclusion

Therefore, the cube of any positive integer is of the form 6q + r, where q is an integer and r = 0, 1, 2, 3, 4, or 5.

Hence proved.

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