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.
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) If a number is divisible by both 2 and 3, then the number is divisible by 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, where 0 ≤ r < 3.All possible remainder = 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 n1 = 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, (n1) and (n+1) is divisible by 3. Then n (n1) (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, (n1) and (n+1) is divisible by 2 then n (n1) (n+1) is also divisible by 2.
Since, n (n1) (n+1) is divisible by both 2 and 3.
Hence, n (n1) (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.