Check weather \[{6^n}\] can end with the digit 0 for any natural number n.

Hint- Prime numbers are generally referred to as the number, which can only be divided by the integer 1 and by itself. In other words, numbers that are greater than 1 but are not the product of smaller numbers. When we factorize a prime number, then they will only have two factors, 1 and the number itself.
The number whose last digit is 0 must be divisible by 2 or 5; hence in this question, check the given number whose power is to be calculated ends with 2 and 5 or not.


Complete step by step solution:
Given n is a natural number for \[{6^n}\], which is the exponential power of 6
A number whose last digit is 0 or the number which ends with 0 should be divisible by 2 and 5, for example:
\[
  10 = 2 \times 5 \\
  100 = 2 \times 2 \times 5 \times 5 \\
 \]
Hence the number ending with digit 0 contains the factors 2 and 5, now check the factorial of
\[6 = 2 \times 3\]
\[{6^n} = {\left( {2 \times 3} \right)^n} = {2^n} \times {3^n}\]
Since 5 is not present in the factorial of 6 hence, we can \[{6^n}\]cannot end with 0.


Note: The natural number is the integers greater than 0; the natural number starts with 1 and increments to infinity 1, 2, 3, 4, 5……..etc. To Check the answer:
\[
  {6^1} = 6 \\
  {6^2} = 36 \\
  {6^3} = 216 \\
  {6^4} = 1296{\text{ and so on}} \\
 \]