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.
SOME OTHER IMPORTANT QUESTIONS
- Use Euclid’s division lemma to show that the cube of any positive integer is of form 9m, 9m+1, and 9m+8.
- State Fundamental theorem of arithmetic.
Related Questions
Prove that for any positive integer n,n3−n is divisible by 6.
How can we prove that n³-n is divisible by 6 if n is any positive integer?
If n is a positive integer, is n³-n divisible by 6?
