what is the units digit of the 99th number in the sequence: 1, 3, 9, 27, ...?

Guest Apr 5, 2017

#3**+1 **

Note that the series can be written as

3^{0}, 3^{1}, 3^{2}, 3^{3}....... =

3^{4n}, 3^{4n + 1}, 3^{4n + 2},^{, }3^{4n + 3}........ for n = 0, 1, 2, 3 ..........

And the last digit has the repeating pattern

1, 3, 9, 7 ...........

So....the 99th term is 3^{98} = 3^{4(24) + 2} which will end in 9

CPhill
Apr 5, 2017