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Question Number 3146 by Rasheed Soomro last updated on 06/Dec/15

Change the order of numbers on a clock−dial so that sum of any two numbers of consecutive positions may be prime.

$$\mathcal{C}{hange}\:{the}\:{order}\:{of}\:{numbers}\:{on}\:{a}\:{clock}−{dial}\:{so}\:{that}\: \\ $$$${sum}\:{of}\:{any}\:{two}\:{numbers}\:\:{of}\:{consecutive}\:{positions}\:{may} \\ $$$${be}\:{prime}. \\ $$

Answered by prakash jain last updated on 06/Dec/15

1→2,4,6,12 2→1,3,5,11 3→2,4,8,10 4→1,3,7,9 5→2,6,8,12 6→1,5,7,11 7→4, 6, 10, 12 8→3, 5,9,11 9→2, 4 10→3, 7, 9 11→2,6,8,12 12→1,5,7,11 9 10 3 2 1 4 7 6 5 12 11 8 9→19 17 10→19 13 3→11 5 2→5 3 1→3 5 4→5 11 7→11 13 6→13 11 5→11 17 12→17 23 11→23 19 8→19 17

$$\mathrm{1}\rightarrow\mathrm{2},\mathrm{4},\mathrm{6},\mathrm{12} \\ $$$$\mathrm{2}\rightarrow\mathrm{1},\mathrm{3},\mathrm{5},\mathrm{11} \\ $$$$\mathrm{3}\rightarrow\mathrm{2},\mathrm{4},\mathrm{8},\mathrm{10} \\ $$$$\mathrm{4}\rightarrow\mathrm{1},\mathrm{3},\mathrm{7},\mathrm{9} \\ $$$$\mathrm{5}\rightarrow\mathrm{2},\mathrm{6},\mathrm{8},\mathrm{12} \\ $$$$\mathrm{6}\rightarrow\mathrm{1},\mathrm{5},\mathrm{7},\mathrm{11} \\ $$$$\mathrm{7}\rightarrow\mathrm{4},\:\mathrm{6},\:\mathrm{10},\:\mathrm{12} \\ $$$$\mathrm{8}\rightarrow\mathrm{3},\:\mathrm{5},\mathrm{9},\mathrm{11} \\ $$$$\mathrm{9}\rightarrow\mathrm{2},\:\mathrm{4} \\ $$$$\mathrm{10}\rightarrow\mathrm{3},\:\mathrm{7},\:\mathrm{9} \\ $$$$\mathrm{11}\rightarrow\mathrm{2},\mathrm{6},\mathrm{8},\mathrm{12} \\ $$$$\mathrm{12}\rightarrow\mathrm{1},\mathrm{5},\mathrm{7},\mathrm{11} \\ $$$$ \\ $$$$\mathrm{9}\:\:\:\mathrm{10}\:\:\mathrm{3}\:\:\mathrm{2}\:\:\mathrm{1}\:\:\mathrm{4}\:\:\mathrm{7}\:\:\mathrm{6}\:\:\mathrm{5}\:\:\mathrm{12}\:\:\mathrm{11}\:\mathrm{8} \\ $$$$\mathrm{9}\rightarrow\mathrm{19}\:\:\:\:\:\mathrm{17} \\ $$$$\mathrm{10}\rightarrow\mathrm{19}\:\:\:\:\:\mathrm{13} \\ $$$$\mathrm{3}\rightarrow\mathrm{11}\:\:\:\:\:\:\mathrm{5} \\ $$$$\mathrm{2}\rightarrow\mathrm{5}\:\:\:\:\:\:\:\:\mathrm{3} \\ $$$$\mathrm{1}\rightarrow\mathrm{3}\:\:\:\:\:\:\:\:\mathrm{5} \\ $$$$\mathrm{4}\rightarrow\mathrm{5}\:\:\:\:\:\:\:\:\mathrm{11} \\ $$$$\mathrm{7}\rightarrow\mathrm{11}\:\:\:\:\:\mathrm{13} \\ $$$$\mathrm{6}\rightarrow\mathrm{13}\:\:\:\:\:\mathrm{11} \\ $$$$\mathrm{5}\rightarrow\mathrm{11}\:\:\:\:\:\mathrm{17} \\ $$$$\mathrm{12}\rightarrow\mathrm{17}\:\:\:\mathrm{23} \\ $$$$\mathrm{11}\rightarrow\mathrm{23}\:\:\:\mathrm{19} \\ $$$$\mathrm{8}\rightarrow\mathrm{19}\:\:\:\:\:\:\mathrm{17} \\ $$

Commented by prakash jain last updated on 06/Dec/15

Thanks. Corrected.

$$\mathrm{Thanks}.\:\mathrm{Corrected}. \\ $$

Commented by RasheedAhmad last updated on 06/Dec/15

V V Nice! In 3^(rd) line I think ′ 1 ′ seems mistakenly included. 3→1_(−) ,2,4,8 ;3+1=4 ∈P

$$\mathcal{V}\:\:\mathcal{V}\:\mathcal{N}{ice}! \\ $$$${In}\:\mathrm{3}^{{rd}} \:{line}\:{I}\:{think}\:'\:\mathrm{1}\:'\:{seems} \\ $$$$\:{mistakenly} \\ $$$${included}. \\ $$$$\mathrm{3}\rightarrow\underset{−} {\mathrm{1}},\mathrm{2},\mathrm{4},\mathrm{8}\:\:\:\:;\mathrm{3}+\mathrm{1}=\mathrm{4}\:\in\mathbb{P} \\ $$

Commented by RasheedAhmad last updated on 06/Dec/15

How many solutions may be?

$$\mathcal{H}{ow}\:{many}\:{solutions}\:{may}\:{be}? \\ $$

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