Question

Show that the cube of any positive integer of yhe form 6q+r q is an integer and r=1,2,3,4,5 is also of the form 6m+r

Answer 1

Answer:

By using division algorithm

By using division algorithma= bq +r

By using division algorithma= bq +r a= positive integer

By using division algorithma= bq +r a= positive integerb = 6

because r = 1,2,3,4,5 and

relationship between r and b is

r<b

q= an integer

Now put the value of a, b, r, q in division algorithm

a=bq+r

a=6q+1 ( here we put value of r =1)

cube both side

3 3

a = ( 6q+1 )

3 3 3 2 2

a = (6 q) + (1) + 3×6q×(1) +3×(6q)×1

3 3 2

a = 216q + 1 + 18q + 108q

3 3 2

a = 216 q + 108q + 18q +1

3 3 2

a = 6 (36q+ 18q + 3 q) +1

3 2

let m= 36q + 18q + 3q

3

a = 6m +1

In above we take value of r = 1 so here we can exchange value of 1 by r

3

a = 6m + r