1-3-2005-have-how-many-digits-in-periodic-part-

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Question Number 403 by 123456 last updated on 25/Jan/15

$$\frac{1}{3^{2005}}\mathrm{have}\mathrm{how}\mathrm{many}\mathrm{digits}\mathrm{in}\mathrm{periodic}\mathrm{part}?$$

Answered by prakash jain last updated on 30/Dec/14

$$3^{2003}\mathrm{digits}\mathrm{will}\mathrm{be}\mathrm{present}\mathrm{in}\mathrm{periodic}\mathrm{part}.$$$${10}^n={(9+1)}^n$$$$n=3^{2003}\mathrm{all}\mathrm{terms}\mathrm{in}\mathrm{binomial}\left(\mathrm{except}1\right)\mathrm{are}$$$$\mathrm{divisible}\mathrm{by}{3}^{2005}.$$$${10}^{3^{2003}}\equiv 1\left(\mathrm{mod}{3}^{2005}\right)$$

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