What is the last digit of the 99th power of 3 multiplied by the 100th power of 7 multiplied by the 101th power of 11?

What is the last digit of the 99th power of 3 multiplied by the 100th power of 7 multiplied by the 101th power of 11?

Because 3*3=9,9*3=27,27*3=81,81*3=243, every four 3s are multiplied, their bits will return to 3, so 99/4=24 other 3s, so the 99th power bit of 3 is the same as 1,7*7=49,49*7=343,343*7=2401,2401*7=16807, so 100/4=25, so the 100th power bit of 7 is 7,11*11=121,121*...

If the absolute value of the square of (a-1)+(b-2)=0, then the 2012 power of a + the third power of b = (-1)+(-1) Squared +(-1) cubic +....+(-1)2012 power +(-1)2013 power result is () If the absolute value of the square of (a-1)+(b-2)=0, then the 2012 power of a + the 3rd power of b = The result of (-1)+(-1) squared +(-1) cubic +....+(-1)2012 power +(-1)2013 power is ()

A-1=0, b-2=0
A=1, b=2
A to the 2012 power + b to the 3rd power
=1+8
=9