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| 1. |
Show that 21n can never end with digits0,4,6 and 8 for any natural number n. |
| Answer» The prime factorisation of 21n\xa0= {tex}( 3 \\times 7 ) ^ { n } = 3 ^ { n } \\times 7 ^ { n }{/tex}Since the prime factorisation of given expression i.e., 21n does not contain 2 and 5, therefore, there is no natural number \'n\'\xa0for which 21n\xa0ends with the digits 0, 2, 4, 6 and 8. | |