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Show that 21 power n cannot end with the digit 0,2,4,6 and 8 for any natural number n |
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Answer» {tex}\\begin{array}{l}21^{\\mathrm n}=3^{\\mathrm n\\;}\\times7^{\\mathrm n}\\\\\\mathrm{Now}\\;3^{\\mathrm n\\;}\\;\\mathrm{and}\\;7^{\\mathrm n\\;}\\;\\mathrm{are}\\;\\mathrm{always}\\;\\mathrm{odd}\\;\\mathrm{for}\\;\\mathrm{all}\\;\\mathrm{values}\\;\\mathrm{of}\\;\\mathrm n\\\\\\mathrm{odd}\\;\\times\\;\\mathrm{odd}\\;=\\mathrm{odd}\\\\\\mathrm{Hence}\\;\\;3^{\\mathrm n\\;}\\times7^{\\mathrm n}\\;\\mathrm{can}\\;\\mathrm{not}\\;\\mathrm{end}\\;\\mathrm{with}\\;2,4,6,8\\end{array}{/tex} It should have an even number as factor in it .but it is not there so it cannot end with digit 0,2,4,6,8 |
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