can any one solve question 4
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as 4^n= (2×2)^n
therefore according to fundamental unit of arithmetic the integer which ends with the digit 0 is of the form (2×5)^n
so 4^n cannot end with the digit 0.
therefore according to fundamental unit of arithmetic the integer which ends with the digit 0 is of the form (2×5)^n
so 4^n cannot end with the digit 0.
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If 4n ends with the digit 0 then it would be divisible by 5. That is the prime factorization of 4n would contain 5 is a prime no. But this is not possible here because the prime factorization of 4n =(2(2))n contain 2 and 2 only as prime factors. So the uniqueness of the fundamental theorm of arithmetic guaranteed that there is no other prime in the factorization of 4n. So there is no natural no. n of which 4n ends with the digit 0.....
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