Problem Solving (PS) | Re: What is the units digit of 4^93*9^15*3^81*7^56*2^9 ?

\(4^{93}*9^{15}*3^{81}*7^{56}*2^9\)

=\(2^{186}*3^{30}*3^{81}*7^{56}*2^9\)

=\(2^{195}*3^{111}*7^{56}\)

Now,\(2^{195} = 2^{4*48+3}\) ; So this will have units digit8

\(3^{111} = 3^ {4*36 + 3}\) will have units digit as 7

\(7^{56} = 7^{4*24}\) will have units digit as 1

Finally we have units digit as 8*7*1 = xx56 , thus Answer must be (E) 6
...

---