**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 lemma, 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.

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** SOME OTHER IMPORTANT QUESTIONS WITH SOLUTION CLICK BELOW QUESTION LINK**

**Prove that if x and y are both odd positive integers then x**^{2 }+y^{2}is even but not divisible by 4.**Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m+1, and 9m+8.**

**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 (n -1) (n +1)**Case I:**When r = 0n = 3q + r = 3q + 0 = 3q is divisible by 3.

**Case II:**When r = 1n = 3q + 1, then n-1 = 3q + 1 – 1 = 3q is divisible by 3.

**Case III:**When r = 2n = 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**Case II:**When r = 1n – 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.

** I hope you have understood this solution**

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