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| 1. |
Show that the number 18 can never end with digit 0 for any natural number N |
| Answer» We have to find out that whether 18n can end with the digit 0 or 5 for any natural number nIf any number ends with the digit 0 then its\xa0factors must be in the form : 2m{tex} \\times {/tex}\xa05nIf\xa0any number ends with the digit 5 then its factors must be in form : 5m {tex} \\times {/tex}\xa03n or 5m{tex} \\times {/tex}\xa07n\xa0So for any number to end\xa0with the digit 0 or 5 its factors must be in form 5m {tex} \\times {/tex}\xa03n or 5m{tex} \\times {/tex}\xa07n\xa0\xa0or\xa02m{tex} \\times {/tex}\xa05n\xa0(where m and n are positive integers)Now, 18 = 2 × 9 = 2 × 3 × 3 = 2\xa0× 32So, 18n\xa0= (2\xa0× 32)n = 2n\xa0× (32)n= 2n\xa0× 32n18n is not in the form of 5m × 3n or 5m × 7n\xa0\xa0or\xa02m × 5n\xa0So, 18n cannot end with 0 or 5 for any natural number n. | |