find last digits of (1023^3923)+(3081^3921)  Select.
A) 8	B) 9	C) 10	D) 11	E) 12

Rationale: we have ( 1023 ^ 3923 ) + ( 3081 ^ 3921 ) ( 3923 / 4 ) = 4 n + 3 ( 3921 / 4 ) = 4 n + 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 4 n + 1 4 n + 2 4 n + 3 4 n 1 1 1 1 1 2 2 4 8 6 3 3 9 7 1 in this way 7 + 1 = 8 is the unit ' s place digit answer : a
Answer: A