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Check whether 6^n can end with the digit 0 for any natural number n . |
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Answer» It never end with the digit 0, as it\'s prime factorization is 2 × 3 = 6. It doesn\'t contain 5, so it can never end with 0. Never because it doesn\'t consist of prime factor as multiple of 2×5 i. e6^n=(2×3)^n |
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