Show that any positive odd integer is of the form (6m+1) or (6m+3)

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.

Show that any positive odd integer is of

 

     SOME OTHER IMPORTANT QUESTIONS WITH SOLUTION CLICK BELOW QUESTION LINK

 

 

    1. Prove that if x and y are both odd positive integers then x+y2 is even but not divisible by 4.
    2. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m+1, and 9m+8.

 

  1. 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 (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 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.

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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.

 

 

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