Solution
Unit's digit of 12 is 2. Now, $$2^1=2$$, $$2^2=4$$, $$2^3=8$$ and $$2^4=16$$ and then again the same cycle is repeated ($$2^5$$ ends in 2).
Thus, numbers of the form $$2^{4n+1}$$ ends in $$2$$
$$2^{4n+2}$$ ends in $$4$$
$$2^{4n+3}$$ ends in $$8$$
$$2^{4n}$$ ends in $$6$$
Now, $$(2)^{123}=(2)^{4n+3}$$
Thus, it must end in 8
=> Ans - (D)
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