Prove that 4^n can never end with digit 0, where n is a natural number.

If the number \(\displaystyle 4^{\mathrm{n}}\), for any n, were to end with digit zero, it would be divisible by $\displaystyle 5$ . So, the prime factorization of \(\displaystyle 4^{\mathrm{n}}\) should contain the prime factor $\displaystyle 5$ . But in \(\displaystyle 4^{\mathrm{n}}=(2 \times 2)^{\mathrm{n}}=2^{2 \mathrm{n}}\), the only prime factor is $\displaystyle 2$ . ∴ By fundamental theorem of arithmetic, there is no natural number n for which \(\displaystyle 4^{\mathrm{n}}\) ends with digit zero.