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My answer is 6. Crude method but still took me a minute.
From condition (1), A is not divisible by 5
for A=1, A^2A=1
for A=3, A^2A=3^6=3^3 * 3^3=27 * 27 => Unit's digit is 9
Hence, cond. (1) is not sufficient
From condition (1), A is even
for A=2, A^2A=2^4=16 => Unit's digit is 6
for A=4, A^2A=4^8= (4^2)^4=16^4 => Unit's digit is 6
for A=6, A^2A=6^12= (6^2)^6=36^6 => Unit's digit is 6
for A=8, A^2A=8^16= (8^2)^8=16^8 => Unit's digit is 6
Because the units digit is 6 before applying the power is 6 and the power is even, it has to give a 6 as the unit's digit in the final answer.
2^4=16, so the unit digit is 6
1^2=1; the units digit is 1, not suff
(2) A is even
2^4=16 --- 6
-2^-4=1/16 not even an integer
10^20 --- 0
Combine: 2^4=16, but -2^4 is not an integer, So it is E for the given verbiage.
If (2) were to be that A is positive and even, then the answer would be C.
Any even= 2N
(2N)^4N= (16*N^4)^N it will always have 6 as its last digits, save that N is a multiple of 5. 0^0 has no sense.
1^4=1 last digit is 1, does not affect 6
2^4=16, does not affect 6
3^4=81, does not affect 6
4^4=256, does not affect 6
5^4=625 DOES AFFECT 6
6^4 never affect 6
7^4=2401 does not affect 6
8^4=4096 does not affect 6
9^4=6561 does not affect 6
and so on