# Show that the cube of any positive integer is of the

Using Euclid’s division Lemma, show that the cube of any positive integer is of form 9m or (9m+1) or (9m+8) for some positive integer m.

Solution: when ‘n’ is divided by 3 to get ‘q’ as a remainder and ‘r’ as a quotient. Applying Euclid’s division lemma,

n = 3q + r, satisfy 0 ≤  r < 3.

All possible remainder = 0, 1, 2

Case I: If r = 0

By Euclid’s division lemma, we get

n = 3 x  q + 0

n = 3 x q

n3 = (3 x q)3

n3 = 27 x q3

n3 = 9 x 3 x q

Let m = 3 x q

n3 = 9m

Case II: If r = 1

By Euclid’s division lemma, we get

n = 3 x q + 1

n3 = (3 x q+1)3

n3 = 27 x q3 + 27 x q2 + 9 x q + 1

n3 = 9 x (3 x q3 + 3 x q2 + 1) + 1

Let m = 3 x q3 + 3 x q2 + 1

n3 = 9m + 1

Case III: If r = 3

By Euclid’s division lemma, we get

n = 3 x q + 2

n3 = (3 x q+2)2

n3 = 27 x q3 + 54 x q2 +36 x q + 8

n3 = 9 x (3 x q3+ 6 x q + 4) + 8

Let m = (3 x q3+ 6 x q + 4)

n3 = 9m + 8

So, the cube of any positive integer is of form 9m or (9m+1) or (9m+8) for some integer m.

Some important question of Real Number

For any positive integer n, prove that n3 – n is divisible by 6.

Solution: when (n3 – n) is divided by 6, then (n3 – n) = n (n -1) (n +1) is also divided by 6.

(n3 – n) = n  x (n -1) x (n +1)

If a number is divisible by both 2 and 3, it will also be divisible by the number 6. When a number (n) is divided by 3, to get q as quotient and r as remainder.

By Euclid’s division lemma, we get

n = 3q + r, 0 ≤ r < 3.

All possible remainders are 0, 1, 2.

Case I: When r = 0

n = 3q + r = 3q + 0 = 3q is divisible by 3.

Case II: When r = 1

n = 3q + 1, then n-1 = 3q + 1 – 1 = 3q is divisible by 3.

Case III: When r = 2

n = 3q + 2, then n+1 = 3q + 2 + 1 = 3q + 3 = 3(q+1) is divisible by 3.

We can say that n, (n-1) and (n+1) is divisible by 3. Then n (n-1) (n+1) is always divisible by 3.

Similarly, when a number is divided by 2 to get q as quotient and r as remainder.

n = 2q + r, where 0  r < 2.

All possible remainder = 0, 1

Case I: when r = 0

n = 2q + 0 = 2q is divisible by 2.

Case II: When r = 1

n = 2q + 1

n – 1 = 2q + 1 – 1 = 2q is divisible by 2.

And n + 1 = 2q + 1 + 1 = 2q + 2 = 2(q+1) is divisible by 2.

We can say that n, (n-1) and (n+1) is divisible by 2 then n (n-1) (n+1) is also divisible by 2.

Since, n (n-1) (n+1) is divisible by both 2 and 3.

Hence, n (n-1) (n+1) = n3 – n is divisible by 6.

Show that any positive odd integer is of the form (6m+1) or (6m+3) or (6m+5), where m is some integer.

Solution:  when ‘n’ is divided by 6 to get ‘m’ as a quotient and ‘r’ be a remainder. Applying Euclid’s division algorithm, we get

n = 6m + r,  0 ≤ r <6

All possible remainder = 0, 1, 2, 3, 4, 5

All possible odd remainder = 1, 3, 5

Case I: when r = 1

By Euclid’s division lemma, we get

n = 6m + 1

In this case, n is clearly odd.

Case II: when r = 3

By Euclid’s division lemma, we get

n = 6m + 3

In this case, n is clearly odd.

Case III: when r = 5

By Euclid’s division lemma, we get

n = 6m + 5

In this case, n is clearly odd.

Hence, any positive odd integer is of the form (6m + 1) or (6m + 3) or 6m + 5), where m is some integer.

Show that the cube of any positive integer is of form 9m or (9m+1) or (9m+8) for some positive integer m.

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My Name is Mukesh Kumar. I am a Teacher, Blogger, Educational Content Writer, and Founder of CBSE Digital Education.

### 6 thoughts on “Show that the cube of any positive integer is of the”

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