A) 3
B) 4
C) 1
D) 2
Correct Answer: A
Solution :
Since, we know that \[{{2}^{2}}\equiv -1(\bmod \,\,5)\] \[(\because \,{{2}^{2}}+1\,\text{divisible by}\,\,5)\] Now, \[{{2}^{2010}}={{({{2}^{2}})}^{1005}}\] \[\equiv {{(-1)}^{1005}}(\bmod \,5)\] \[\Rightarrow \] \[{{2}^{2010}}\equiv -1(\bmod 5)\] \[\Rightarrow \] \[-1\equiv {{2}^{2010}}(\bmod \,5)\] But \[{{2}^{2010}}\equiv 3x(\bmod \,5)\] \[\Rightarrow \] \[-1\equiv 3x(\bmod 5)\] (by transitive relation) \[\Rightarrow \] \[x=3\]
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