Consider the numbers 4^{n},where n is a natur | vedclass

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(C) If the number $4^{n}$ for any natural number $n$ were to end with the digit zero,it would be divisible by $5$. This implies that the prime factori

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No,there is no such value of $n$

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(C) If the number $4^{n}$ for any natural number $n$ were to end with the digit zero,it would be divisible by $5$. This implies that the prime factorization of $4^{n}$ must contain the prime number $5$. However,we can write $4^{n} = (2^{2})^{n} = 2^{2n}$. The only prime factor in the prime factorization of $4^{n}$ is $2$. According to the Fundamental Theorem of Arithmetic,this prime factorization is unique. Since $5$ is not a factor of $4^{n}$,there is no natural number $n$ for which $4^{n}$ ends with the digit zero.

Consider the numbers $4^{n}$,where $n$ is a natural number. Check whether there is any value of $n$ for which $4^{n}$ ends with the digit zero.

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