** 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

n^{3} = (3 x q)^{3}

^{ }n^{3} = 27 x q^{3}

^{ }n^{3} = 9 x 3 x q

Let m = 3 x q

^{ }n^{3} = 9m

** Case II: **If r = 1

** **By Euclid’s division lemma, we get

n = 3 x q + 1

n^{3} = (3 x q+1)^{3}

n^{3} = 27 x q^{3} + 27 x q^{2} + 9 x q + 1

n^{3} = 9 x (3 x q^{3} + 3 x q^{2} + 1) + 1

Let m = 3 x q^{3} + 3 x q^{2} + 1

n^{3} = 9m + 1

** Case III: **If r = 3

** **By Euclid’s division lemma, we get

n = 3 x q + 2

n^{3 }= (3 x q+2)^{2}

n^{3 }= 27 x q^{3} + 54 x q^{2} +36 x q + 8

n^{3} = 9 x (3 x q^{3}+ 6 x q + 4) + 8

Let m = (3 x q^{3}+ 6 x q + 4)

n^{3} = 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 n ^{3} – n is divisible by 6.**

** Solution: **when (n^{3} – n) is divided by 6, then (n^{3} – n) = n (n -1) (n +1) is also divided by 6.

(n^{3} – 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) = n^{3} – 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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