Menu Close

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




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

Leave a Reply

Your email address will not be published. Required fields are marked *