Show that 9" cannot end with digit 2 for any ne N.
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We can see prime factorization on 9^ndoes not contain 2. Hence it can't endwith 2. to end with number 2, 9n showdivisible by 2 and the prime factorisation of 9n should cannot the number 2 but the prime factorisation of 9n will be 1 and 3 thus cannot end with digit 2 for any natural number
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We can see prime factorization on 9^n does not contain 2. Hence it can't end with 2. to end with number 2, 9n show divisible by 2 and the prime factorisation of 9n should cannot the number 2 but the prime factorisation of 9n will be 1 and 3 thus cannot end with digit 2 for any natural number.
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