Show that any number of the form ` 6^(n)` , where ` n ne N` can never and with the digit 0.

Explanation

If `6^(n)` end with 0 then it must have 5 as a factor.
Butm, ` 6^(n)= ( 2 xx 3)^(n) = ( 2^(n) xx 3^(n) )` shows that 2 and 4 are the only prime factors of ` 6^(n)`.
Also, we know form the fundamental theorem of arithmetic that the prime factorisation of each number is unique.
SO, is not a factor of ` 6 ^(n)`
Hence,` 6^(n)` can not a factor of ` 6^(n)` .
Hence,` 6^(n)` can never end with the digit 0.