Question

Check weather $$6^n$$ can end with the digit 0 for any natural number n.

Answer 1

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:
$$\begin{gathered} 10=2\times 5 \\ 100=2\times 2\times 5\times 5 \end{gathered}$$
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:
$$\begin{gathered} 6^1=6 \\ {6}^2=36 \\ {6}^3=216 \\ {6}^4=1296\text{ and so on} \end{gathered}$$