prove that there is no natural number for which 4^n ends with the digit 0
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If it ends by 0 then it prime factorisation must contain factor 5 or 0
4^n=(2×2)^n
there in no factor 5 or 0 in the prime factorisation of 4^n.
hence proved
4^n=(2×2)^n
there in no factor 5 or 0 in the prime factorisation of 4^n.
hence proved
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