Question

prove that there is no natural number n for which 4ñ ends with the digit 0

Answer 1

If the digit 4^n were to end with the digit 0 it should be divisble by 5 i.e. the prime factorization of 4^n would contain the prime 5. This is not possible because 4^n =(2)^2n. Hence the prime factorization of 4^n only contains the factor 2. Hence the uniqueness of the fundamental theorem of arithmetic guarantees that there are no other primes in the factorization of (4)^n except 2. Hence, There is no natural no. n for which 4^n ends with the digit zero..

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