Real Numbers

Question

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

Let a and b be two positive integergs such that a is greater than b; then :

a = bq + r;

where q and r are positive integers 0 ≤ r < b.

Taking b = 3, we get

a = 3q + r ; where 0 ≤ r < 3.

⇒ Different values of integer a are 3q, 3q + 1 or 3q + 2.

Cube of 3q = (3q)3

= 27q3 = 9(3q3) = 9m ;

where m is some integer.

Cube of 3q + 1 = (3q + 1)3

= (3q + 3(3q)2 × 1 + 3(3q) × 12+ l3

[∵ (a + b)3 = a3 + 3a2b + 3ab2 + 1]

= 27q3 + 27q2 + 9q + 1

= 9(3q3 + 3q2 + q) + 1

= 9m + 1; where m is some integer.

Cube of 3q + 2 = (3q + 2)3

= (3q)3 + 3(3q)2 × 2 + 3 × 3q × 22 + 23

= 27q3 + 54q2 + 36q + 8

= 9(3q3 + 6q2 + 4q) + 8 = 9m + 8; where m is some integer.

∴ Cube of any positive integer is of the form 9m or 9m + 1 or 9m + 8.

Sponsor Area