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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