Consider a number 4^n, where n is a natural number . Check whether there is any value of n for which 4^n ends with0
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let n be any natural number 1,2,3..........
(4)1=4
(4)2=16
(4)3=64
(4)4=256.... therefore 4^n never ends with 0 .and as it end with 6and4 thatswhy it is divisible by 2
(4)1=4
(4)2=16
(4)3=64
(4)4=256.... therefore 4^n never ends with 0 .and as it end with 6and4 thatswhy it is divisible by 2
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