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Let n=3^(100) then for n the value of te...

Let `n=3^(100)` then for n the value of ten's digit is

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`n = 3^100 = (3^2)^50 = (10-1)^50`
`:. n = C(50,0)10^50 (-1)^0+ C(50,1)10^49 (-1)^1+...+C(50,48)10^2 (-1)^48+C(50,49)10^1 (-1)^49+C(50,50)10^0 (-1)^50`
Now, value of ten`'`s digit will depend on last three terms of the above expression as initial terms will contain many zeroes.
`:. n = ...+C(50,48)10^2 (-1)^48+C(50,49)10^1 (-1)^49+C(50,50)10^0 (-1)^50`

`:. n = ...+25*49*100 -50*10 +1`
`=> n = ...+122500 -500 +1`
`=>n = ...+122001`
So, ten`'`s will be `0` and last two digits will be `01`.
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