Chapter 1: Real Numbers

6q + r is a positive integer, where q is an integer and r = 0, 1, 2, 3, 4, 5

Then, the positive integers are of form 6q, 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5.

Taking cube on L.H.S and R.H.S,

For 6q,

(6q)³ = 216 q³ = 6(36q)³ + 0

= 6m + 0, (where m is an integer = (36q)³)

For 6q+1,

(6q+1)³ = 216q³ + 108q² + 18q + 1

= 6(36q³ + 18q² + 3q) + 1

= 6m + 1, (where m is an integer = 36q³ + 18q² + 3q)

For 6q+2,

(6q+2)³ = 216q³ + 216q² + 72q + 8

= 6(36q³ + 36q² + 12q + 1) +2

= 6m + 2, (where m is an integer = 36q³ + 36q² + 12q + 1)

For 6q+3,

(6q+3)³ = 216q³ + 324q² + 162q + 27

= 6(36q³ + 54q² + 27q + 4) + 3

= 6m + 3, (where m is an integer = 36q³ + 54q² + 27q + 4)

For 6q+4,

(6q+4)³ = 216q³ + 432q² + 288q + 64

= 6(36q³ + 72q² + 48q + 10) + 4

= 6m + 4, (where m is an integer = 36q³ + 72q² + 48q + 10)

For 6q+5,

(6q+5)³ = 216q³ + 540q² + 450q + 125

= 6(36q³ + 90q² + 75q + 20) + 5

= 6m + 5, (where m is an integer = 36q³ + 90q² + 75q + 20)

Hence, the cube of a positive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r.

According to Euclid’s division Lemma,

Let the positive integer = n

And b=3

n =3q+r, where q is the quotient and r is the remainder

0<r<3 implies remainders may be 0, 1 and 2

Therefore, n may be in the form of 3q, 3q+1, 3q+2

When n=3q

n+2=3q+2

n+4=3q+4

Here n is only divisible by 3

When n = 3q+1

n+2=3q=3

n+4=3q+5

Here only n+2 is divisible by 3

When n=3q+2

n+2=3q+4

n+4=3q+2+4=3q+6

Here only n+4 is divisible by 3

So, we can conclude that one and only one out of n, n + 2 and n + 4 is divisible by 3.

Hence Proved