Last digit on $3^{100}$ How to find the last digit on $3^{100}$?
Is there any proper method to solve such questions without calculator of course?

$3^{100}=(3^4)^{25 }=81^{25}=(80 + 1)^{25}=a*80^{25}+b*80^{24}+.....+x*80^2+y*80 + 1$  where a,b,c,....,y are integer coeffients obtained by expanding $(80 +1)^{25} $.  We can calculate these coefficients if we wanted to, but it's enough to know they are integers.  
Notice, every term $k*80^j $ is divisible by 10 so
$3^{100}=10M + 1$ where $M= (a*80^{25}+.....+y*80)/10$.  $M $ is a whole number, so the last digit is 1.
This is more detail than we need.
It's a basic fact that if we use the notation $a \equiv b \mod n $ to mean that for integers $a,b$ and $n $ that $a$ and $b $ will have the same remainder when divided by $n $ that if $a \equiv 1 \mod n $ then $ak \equiv k \mod n $.
(Because $a \equiv 1 \mod n$ means $a=jn +1$ so $ak =(jk)n+k $ so $ak \equiv k \mod n$.)
Therefore if $a^k \equiv 1 \mod n $ than $a^{mk}=a^k*a^k*....a^k\equiv 1*1*...*1 \mod n $.
So as $3^4=81 \equiv 1 \mod 10$, it follows that $3^{4m} \equiv 1 \mod 10$ and will have 1 as a last digit.
More interesting is what are the last 2 digits of $3^{100} $.  $3^4=81 \mod 100$ so $3^8=6400+160+1\equiv 61 \mod 100$.  $3^{12} \equiv 81*61 \equiv (6+8)*10+1\equiv 41 \mod 100$.  $3^{16}\equiv 61^2\equiv (6+6)10+1\equiv 21\mod 100$.  $3^{20}\equiv 21*81\equiv (8+2)10+1 \equiv 1701 \equiv 1\mod 100$.  So $3^{20m}\equiv 1 \mod 100$ so the last two digits of $3^{100}$ are 01.
