find last digits of (1023^3923)+(3081^3921)  Choose the correct choice from the following options.
A) 8	B) 9	C) 10	D) 11	E) 12
we have (1023^3923)+(3081^3921)   (3923/4)=4n+3 (3921/4)=4n+1 where n can be any integer   now through cyclicity we can obtain or you can learn the chart from arun sharma   number ending in 4n+1 4n+2 4n+3 4n   1 1 1 1 1   2 2 4 8 6   3 3 9 7 1   in this way 7+1= 
<gadget id="calculator">7+1</gadget>
<output>8</output>
8 is the unit's place digit   8NSWER:8
<result>A</result>
find last digits of (1023^3923)+(3081^3921)