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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