NIELIT 2021 Dec Scientist B - Section B: 87
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Convert $(503201)_{6}$ into $(?)_{4}$

  1. $12122231$
  2. $11000011$
  3. $21222301$
  4. $22323301$
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$(503201)_6\rightarrow(39601)_{10}\rightarrow(21222301)_4$
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$(503201)_6\rightarrow(39601)_{10}\rightarrow(21222301)_4$

Base $6$ to base $10$ conversion:

$(1*6^0+0*6^1+2*6^2+3*6^3+0*6^4+5*6^5=(39601)_{10}$

Now base $10$ to base $4$ conversion:

 

Option $(C)$ is correct.

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