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Question Number 403 by 123456 last updated on 25/Jan/15
(1/3^(2005) ) have how many digits in periodic part?
$$\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2005}} }\:\mathrm{have}\:\mathrm{how}\:\mathrm{many}\:\mathrm{digits}\:\mathrm{in}\:\mathrm{periodic}\:\mathrm{part}? \\ $$
Answered by prakash jain last updated on 30/Dec/14
3^(2003) digits will be present in periodic part. 10^n =(9+1)^n n=3^(2003) all terms in binomial (except 1) are divisible by 3^(2005) . 10^3^(2003) ≡1(mod 3^(2005) )
$$\mathrm{3}^{\mathrm{2003}} \:\mathrm{digits}\:\mathrm{will}\:\mathrm{be}\:\mathrm{present}\:\mathrm{in}\:\mathrm{periodic}\:\mathrm{part}. \\ $$$$\mathrm{10}^{{n}} =\left(\mathrm{9}+\mathrm{1}\right)^{{n}} \\ $$$${n}=\mathrm{3}^{\mathrm{2003}} \:\mathrm{all}\:\mathrm{terms}\:\mathrm{in}\:\mathrm{binomial}\:\left(\mathrm{except}\:\mathrm{1}\right)\:\mathrm{are} \\ $$$$\mathrm{divisible}\:\mathrm{by}\:\mathrm{3}^{\mathrm{2005}} . \\ $$$$\mathrm{10}^{\mathrm{3}^{\mathrm{2003}} } \equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{3}^{\mathrm{2005}} \right) \\ $$

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