# show that the cube of any positive integer is of the form 9m,9m+1 or 9m+8

Let a be any positive integer and b = 3
a = 3q + r, where q ≥ 0 and 0 ≤ r < 3

Therefore, every number can be represented as these three forms. There are three cases.
Case 1: When a = 3q,

Where m is an integer such that m =
Case 2: When a = 3q + 1,
a 3 = (3q +1) 3
a 3 = 27q 3 + 27q 2 + 9q + 1
a 3 = 9(3q 3 + 3q 2 + q) + 1
a 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 + 8
a 3 = 9(3q 3 + 6q 2 + 4q) + 8
a 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.

• 167

euclid's lemma states that a=bq +r

let 'a' be any positive integer, 'b' =3 and quotient ='q'.

r=0,1,2

when r=0, a=3q +0 =3q

a3=[3q]3= 27q3 = 9[3q3] = 9m where m=[3q3]

when r=1, a=3q+1

a3 = [3q+1]3

= [3q]3 +[1]3 +3[3q][1]{3q+1}

= 27q3 +27q2 + 9q + 1

=9{3q3 + 3q2 + q} + 1  =9m+1 where 'm' =3q3 + 3q2 + q

when r=2, a= 3q + 2

a3=[3q+2]3

=27q3 + 54q2 + 36q + 8

=9{3q3+ 6q2+4q} +8   =9m +8 hwre 'm' = 3q3+6q2+4q

hence proved

• 22

a=mq+r (0 < r)

Here, b=9, Then, a=9m+r(q > 0)

Now, r=o or 9 (0< r<9)

This means that the remainder can either be 0,1,2,3,4,5,6,7 or 8

That is,(9m+1),(9m+2) and so on.

HOPE THIS HELPS YOU .....

• -22
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