A) \[1\]
B) \[2\]
C) \[3\]
D) \[4\]
Correct Answer: D
Solution :
Now, \[{{3}^{2}}\equiv 4\,\,(\bmod \,5)\] \[\Rightarrow \] \[{{({{3}^{2}})}^{2}}\equiv 16\,(\bmod \,5)\equiv 1\,(\bmod 5)\] \[\Rightarrow \] \[{{({{3}^{4}})}^{25}}\equiv {{(1)}^{25}}(\bmod 5)\] \[\Rightarrow \] \[{{3}^{100}}\equiv 1\,(\bmod 5)\] and \[{{2}^{2}}\equiv (\bmod \,5)\] \[\Rightarrow \] \[{{({{2}^{2}})}^{2}}\equiv 16(\bmod \,5)\equiv 1(\bmod \,5)\] \[\Rightarrow \] \[{{({{2}^{4}})}^{12}}\equiv {{1}^{12}}(\bmod \,5)\equiv 1(\bmod \,5)\] \[\Rightarrow \] \[{{2}^{48}}{{.2}^{2}}\equiv 4(\bmod 5)\] \[\therefore \] \[{{3}^{100}}\times {{2}^{50}}=1\times 4(\bmod 5)\] Hence, remainder is 4.
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