Solution:

If the number 6n closures with the digit zero (0), then, at that point it ought to be distinct by 5, as we probably are aware any number with unit place as 0 or 5 is detachable by 5.

Prime factorization of 6n = (2×3)n

Consequently, the great factorization of 6n doesn’t contain indivisible number 5.

Henceforth, unmistakably for any regular number n, 6n isn’t distinguishable by 5 and along these lines it demonstrates that 6n can’t end with the digit 0 for any normal number n.