Question

Using euclid's division lemma show that the cube of any positive integer is of the form 9m 9m + 199 + 8

Answer 1

According to Euclids Division Lemma

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

Then using Euclids 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,

So 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 get9m +1

b = 2 then 2³ = 8 so we get 9m + 8

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

b = 4then 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.

Answer 2

Step-by-step explanation:

Let a be any positive integer and b = 3

a = 3q + r, where q ≥ 0 and 0 ≤ r < 3

∴ r = 0,1,2 .

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 = (3q +1) ³

a = 27q ³+ 27q ² + 9q + 1

a = 9(3q ³ + 3q ² + q) + 1

a = 9m + 1 [ Where m = 3q³ + 3q² + q ) .

Case 3: When a = 3q + 2,

a = (3q +2) ³

a = 27q³ + 54q² + 36q + 8

a = 9(3q³ + 6q² + 4q) + 8

a = 9m + 8

Where m is an integer such that m = (3q³ + 6q² + 4q)

Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

Hence, it is proved .

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THANKS.