Show that 21^n can not end with the digit 0 for any natural number n.
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If any of the numbers 21^n ends in 0,2,4,6,8 that means that 21^n has 2 in its prime factorization. But 21^n = 3^n * 7^n and each number can have only one unique prime factorization. So there can't be 2 in the prime factorization of 21^n
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