the power of any number is a multiple of itself let us give the values of n 6,6×6,6×6×6…. 6 = 2 × 3 6 contains only two factors 2 and 3 but a digit to end with zero we require 5 as a factor hence 6 to the power of n cannot end with the digit 0 for any natural number n Reply
. It can be observed that 5 is not in the prime factorisation of 6n. Hence, for any value of n, 6n will not be divisible by 5. Therefore, 6n cannot end with the digit 0 for any natural number n. Reply
the power of any number is a multiple of itself
let us give the values of n
6,6×6,6×6×6…. 6 = 2 × 3
6 contains only two factors 2 and 3
but a digit to end with zero we require 5 as a factor
hence 6 to the power of n cannot end with the digit 0 for any natural number n
. It can be observed that 5 is not in the prime factorisation of 6n. Hence, for any value of n, 6n will not be divisible by 5. Therefore, 6n cannot end with the digit 0 for any natural number n.