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Question

When $$\displaystyle 102^{29}$$ is divided by $$5$$ we get remainder


A
2
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B
3
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C
4
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D
1
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Solution

The correct option is C $$2$$
$$N={ (2\times 51) }^{ 29 }$$
      $$={ 2 }^{ 29 }.{ (50+1) }^{ 29 }$$
      $$= { 2 }^{ 29 }(50\lambda +1)$$            $$;\lambda \in I$$
      $$ = { 2 }^{ 29 }50\lambda +{ 2 }^{ 29 }$$
      $$= 50{ \lambda  }^{ 1 }+{ { 2.(2 }^{ 2 }) }^{ 14 }$$       $${ ;\lambda  }^{ 1 }\in I$$
      $$= 50{ \lambda  }^{ 1 }+2.(5-{ 1) }^{ 14 }$$
      $$ = 50{ \lambda  }^{ 1 }+2.(5k+{ 1) }$$
      $$= 50{ \lambda  }^{ 1 }+10K+2$$
      $$= 5(10{ \lambda  }^{ 1 }+2K)+2$$
$$\therefore$$  In dividing by $$5$$; we get the remainder $$ =2.$$
Hence the answer is $$2.$$


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