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Question Number 26686 by Rasheed.Sindhi last updated on 28/Dec/17

Commented by Rasheed.Sindhi last updated on 28/Dec/17

$$\text{You can't use 'macro parameter character \#' in math mode}$$$$\mathrm{reposted}\mathrm{here}\mathrm{to}\mathrm{refocus}\mathrm{the}\mathrm{question}.$$

Answered by Rasheed.Sindhi last updated on 28/Dec/17

$$\mathrm{The}\mathrm{bigger}\mathrm{circle}\mathrm{is}\mathrm{divided}\mathrm{into}$$$$\mathrm{quarters}.\mathrm{Each}\mathrm{quarter}\mathrm{contains}$$$$\mathrm{three}\mathrm{numbers}:\mathrm{Outer}\mathrm{number}\left(\mathrm{O}\right)$$$$\mathrm{and}\mathrm{two}\mathrm{inner}\mathrm{numbers}\left({\mathrm{i}}_1\mathrm{\&}{\mathrm{i}}_2\right).$$$$\mathrm{The}\mathrm{relation},{\mathrm{I}}^{\prime }\mathrm{ve}\mathrm{discovered}\mathrm{is}:$$$$$$$$\left(\mathrm{O}\right)\mathrm{mod}({\mathrm{i}}_1+{\mathrm{i}}_2)\mathrm{is}\mathrm{whole}\mathrm{power}\mathrm{of}2$$$$65\mathrm{mod}(2+7)=2=2^1$$$$233\mathrm{mod}(6+9)=8=2^3$$$$130\mathrm{mod}(3+4)=4=2^2$$$$300\mathrm{mod}(5+8)=1=2^0$$$$\mathrm{But}$$$$141\mathrm{mod}(5+8)={11}^{\ast }$$$$278\mathrm{mod}(5+8)=5^{\ast }$$$$306\mathrm{mod}(5+8)=7^{\ast }$$$$\ast 11,5\mathrm{\&}7\mathrm{are}\mathrm{not}\mathrm{whole}\mathrm{powers}$$$$\mathrm{of}2$$$$\therefore ?=300$$

Commented by shivram198922@gmail.com last updated on 30/Dec/17

$$thanks$$

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