Question

Use euclid's division lemma to show that the cube of any positive integer is of the form 9q or 9q+8, where q is some integer.

Answer 1

According to Euclid’s Division Lemma

Let take a as any positive integer and b = 9.

Then using Euclid’s algorithm we get a = 9q + r here r is remainder and value of q is more than or equal to 0 and r = 0, 1, 2, 3, 4, 5 , 6 , 7 , 8, because 0 ≤r < b and the value of b is 9

Sp possible forms will 9q, 9q+1, 9q+2,9q+3,9q+4,9q+5,9q+6,9q+7 and 9q+8

to get the cube of these values use the formula

(a+b)³ = a³ + 3a²b+ 3ab² + b³

In this formula value of a is always 9q

So plug the value we get

(9q+b)³ = 729q³ + 243q²b + 27qb² + b³

Now divide by 9 we get quotient = 81q³ + 27q²b + 3qb² and remainder is b³

So we have to consider the value of b³

b = 0 we get 9m+0 = 9m

b = 1 then 1³ = 1 so we get 9m +1

b = 2 then 2³ = 8 so we get 9m + 8

b = 3 then 3³ = 27 and it is divisible by 9 so we get 9m

b = 4 then 4³ =64 divide by 9 we get 1 as remainder so we get 9m +1

b=5 then 5³=125 divide by 9 we get 8 as remainder so we get 9m+8

b=6 then 6³=216 divide by 9 no remainder there so we get 9m

b=7 then 7³ = 343 divide by 9 we get 1 as remainder so we get 9m+1

b=8 then 8³ = 512 divide by 9 we get 8 as remainder so we get 9m+8

So all values are in form of 9m , 9m+1 or 9m+8