Solution : 3^27= ((3^4)^6) * (3^3) = (81^6) * 27 then unit digit of (81^6) is 1 so on multiplying with 27,
unit digit in the result will be 7. now, 7 when divided by 5 gives 2 as remainder.

I get the first part but the unit digit aspect is confusing, can anyone explain

Vineet said: 8/10/2014

i think given explanation is slightly confusing.
So, my explanation may help you in understanding the answer :-
3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = 243 (unit digit is 3)
3^6 = unit digit is 9....
So, unit digit is repeating in this order - 3, 9, 7, 1, 3, 9, 7, 1......So, in this way for 3^27, unit digit will be 7. and for divisibility by 5 unit digit is either 0 or 5. so 3^27 will have unit digit 7 and when divided by 5 remainder will be 2.

samar said: 8/30/2014

I am confused.

lijesh said: 9/18/2014

thank u vineet ,,,,,

sree said: 9/29/2014

am not understanding....please explain

Saravanan said: 10/4/2014

In the number 81 (3^4), the unit digit is 1 and tenth digit is 8. If you multiply (81^6 - 81*81*81*81*81*81) the unit digit will always be 1. Therefore, if you multiply 27 with the number that has unit digit as 1, the unit digit of a product must be 7. Finally, if you divide any number that has 7 as a unit digit by 5 the remainder always be 2.

Tejal said: 11/22/2014

Writing too late.. just in case anyone needs explanation for the calculation
3^27 = (3^24) + (3^3)
= (3^(4*6)) + (3^3)
= ((3^4)^6)+(3^3)
=((3^4)^6)+27
=(81^6)+27
the rest of the solution is evident