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Prove that $4^n$ can never end with digit $0$, where $n$ is a natural number.
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If the number $4^n$, for any $n$, were to end with digit zero, it would be divisible by $5$. So, the prime factorization of $4^n$ should contain the prime factor $5$.
But in $4^n = (2 \times 2)^n = 2^{2n}$, the only prime factor is $2$.
$\therefore$ By fundamental theorem of arithmetic, there is no natural number $n$ for which $4^n$ ends with digit zero.
But in $4^n = (2 \times 2)^n = 2^{2n}$, the only prime factor is $2$.
$\therefore$ By fundamental theorem of arithmetic, there is no natural number $n$ for which $4^n$ ends with digit zero.