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

Step-by-step solution

1. A number ends with the digit $0$ only if it is divisible by $10$.
2. Since $10=2\times 5$, such a number must have both $2$ and $5$ in its prime factorisation.
3. Now factorise $6^n$. Since $6=2\times 3$, we get $6^n=2^n\times 3^n$.
4. The only primes appearing here are $2$ and $3$. There is no factor of $5$.
5. By the Fundamental Theorem of Arithmetic the prime factorisation of a number is unique, so a factor of $5$ cannot appear from anywhere else.
6. Since $5$ never divides $6^n$, neither does $10$, and so $6^n$ cannot end in $0$ for any natural number $n$.

Concept tested: Fundamental Theorem of Arithmetic

Related questions

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