Question

Question 5
Use Euclid's division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Answer 1

Let 'a' be any positive integer and b = 3.
a = 3q + r, where q ≥ 0 and 0 ≤ r < 3
$\therefore$ a = 3q or 3q + 1 or 3q + 2
Therefore, every number can be represented as these three forms. There are three cases.
Case 1: When a = 3q,
$a^3=(3q{)}^3=27q^3=9(3q{)}^3=9m,$
Where m is an integer such that m = $(3q{)}^3$
Case 2: When a = 3q + 1,
$a^3=(3q+1{)}^3{a}^3=27q^3+27q^2+9q+1a^3=9(3q^3+3q^2+q)+1a^3=9m+1$
Where m is an integer such that m = $(3q^3+3q^2+q)$
Case 3: When a = 3q + 2,
$a^3=(3q+2{)}^3{a}^3=27q^3+54q^2+36q+8a^3=9(3q^3+6q^2+4q)+8a^3=9m+8$
Where m is an integer such that m = $(3q^3+6q^2+4q)$
Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.