A) 4
B) 8
C) 2
D) 6
Correct Answer: C
Solution :
Since \[{{2}^{1}}=2,\,\,{{2}^{2}}=4,\,{{2}^{3}}=8,\,{{2}^{4}}=16,\,{{2}^{5}}=32\] It is clear that unit place is repeated after every four power then, \[{{2}^{301}}={{({{2}^{4}})}^{75}}.2\] \[={{\text{(}16\text{)}}^{75}}.2\] \[\therefore \] digit at unit place in \[{{(16)}^{75}}\] is 6 \[\therefore \] Digit at units place in \[{{2}^{301}}=\] digit at units place in (6). 2 = 2 Hence die remainder, when \[{{2}^{301}}\] is divided by 5, is 2.
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