show that n3-n is divisible by 3 for any positive integers n
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= n^3 -n
= n(n^2-1)
= n(n^2 -1^2)
= n(n+1)(n-1)
We know if sum of digits is divisible by 3 , no. is divisible by 3 .
= n+n+1+n-1
= 3n (divisible by 3)
Hence , the n^3-n is divisible by 3 for any positive integer n .
= n^3 -n
= n(n^2-1)
= n(n^2 -1^2)
= n(n+1)(n-1)
We know if sum of digits is divisible by 3 , no. is divisible by 3 .
= n+n+1+n-1
= 3n (divisible by 3)
Hence , the n^3-n is divisible by 3 for any positive integer n .
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