Is x divisible by 30?

B

\(30=2*3*5\)

1. \(x = k(m^3 - m)=k*m*(m^2-1)=k*m*(m-1)(m+1)\)

\((m-1),m,(m+1)\) are consecutive integers and the product of these integers will always be divisible by 6 (2 and 3).
But the consequence can be divisible by 5 or cannot be. Therefore, if k and the product are not divisible by 5, x will not be divisible by 30; if k or the product is divisible by 5, x will be divisible by 30. INSUFF.

2. \(x = n^5 - n=n*(n^4-1)=n*(n^2-1)*(n^2+1)=n*(n-1)*(n+1)*(n^2+1)\)

\((n-1),n,(n+1)\) are consecutive integers and the product of these integers will always be divisible by 6 (2 and 3).

if n=5k, 5k+1, or 5k+4, the product will be divisible by 5 and therefore, by 30.
if n=5k+2 or 5k+3 the product will not be divisible by 5.
but: n=5k+2: \((n^2+1)=(5k+2)^2+1=25k^2+20k+5=5*(5k^2+4k+1)\) Therefore, x is divisible by 30
n=5k+3: \((n^2+1)=(5k+3)^2+1=25k^2+30k+10=5*(5k^2+6k+2)\) Therefore, x is also divisible by 30.
SUFF.