Question-26686

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

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

Question#26312 by Mr Shivram198922 reposted here to refocus the question.

$$\mathrm{Question}#\mathrm{26312}\:\mathrm{by}\:\mathrm{Mr}\:\mathrm{Shivram198922} \\ $$$$\mathrm{reposted}\:\mathrm{here}\:\mathrm{to}\:\mathrm{refocus}\:\mathrm{the}\:\mathrm{question}. \\ $$

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

The bigger circle is divided into quarters. Each quarter contains three numbers: Outer number(O) and two inner numbers(i_1 & i_2 ). The relation, I′ve discovered is: (O)mod(i_1 +i_2 ) is whole power of 2 65 mod(2+7)=2=2^1 233 mod(6+9)=8=2^3 130 mod(3+4)=4=2^2 300 mod(5+8)=1=2^0 But 141 mod(5+8)=11^∗ 278 mod(5+8)=5^∗ 306 mod(5+8)=7^∗ ∗ 11,5 & 7 are not whole powers of 2 ∴ ? = 300

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

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

$${thanks} \\ $$

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