Show that 9^n cannot end with digit 3
for any natural number n
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Answered by
2
9×9 = 81
9×9×9= 729
9×9×9×9= 6561
9×9×9×9×9=59049
its last digit is ending either 1 or 9
not 3......
9×9×9= 729
9×9×9×9= 6561
9×9×9×9×9=59049
its last digit is ending either 1 or 9
not 3......
Answered by
0
9×9 = 81
9×9×9= 729
9×9×9×9= 6561
The last digit is not ending with 3.
Hence proved.
9×9×9= 729
9×9×9×9= 6561
The last digit is not ending with 3.
Hence proved.
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