A) \[7\]
B) \[3\]
C) \[1\]
D) \[9\]
Correct Answer: D
Solution :
The digit in the unit place of (2009) is 0. Now, \[{{3}^{1}}=3,\] \[{{3}^{2}}=9,\] \[{{3}^{3}}=27,\] \[{{3}^{4}}=81,\] \[{{3}^{5}}=243\] \[\therefore \] \[{{3}^{7886}}={{(3)}^{4}}^{1971}{{3}^{2}}\] The digit in the unit place of \[{{3}^{7886}}\] is 9. \[\therefore \] The digit in the unit place of \[\left( 2009 \right)!+{{3}^{7886}}\]is 9.You need to login to perform this action.
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