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

Solution:

If the number $6^{\mathrm{n}}$, for any natural number $\mathrm{n}$, ends with digit 0 , then it would be divisible by 5 . That is, the prime factorisation of $6^{\mathrm{n}}$ would contain the prime number 5 . This is not possible because $6^{\mathrm{n}}=(2 \times 3)^{\mathrm{n}}=2^{\mathrm{n}} \times 3^{\mathrm{n}} ;$ so the only primes in the factorisation of $6^{\mathrm{n}}$ are 2 and 3 and the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of $6^{\mathrm{n}}$. So, there is no natural number $\mathrm{n}$ for which $6^{\mathrm{n}}$ ends with the digit zero.