Solution:
If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5
It can be observed that 5 is not in the prime factorisation of (22x7)n or 28n.
Hence, for any value of n, 28n will not be divisible by 5.
Therefore, 28n cannot end with the digit 0 for any natural number n.
