jasen 4 Posted January 26, 2016 Report Share Posted January 26, 2016 1. Easy Puzzle Put numbers 1,2,3,5,6,7 to each point, so : I. The sum of points in every circle are equal, (A+E+F+C = B+E+D+C = A+D+F+B) II. A+B+C = D+E+F There are 2 solutions. 2 .Harder Puzzle Find the value of x, then put number 2,3,4,9,12,x to each point, so : I. The product of points in every circle are equal, (A*E*F*C = B*E*D*C = A*D*F*B) II. A*B*C = D*E*F There are 2 solutions. Quote Link to post Share on other sites

0 Solution DejMar 9 Posted January 26, 2016 Solution Report Share Posted January 26, 2016 Spoiler 1. As the total of (1+2+3+5+6+7)/2 = 12, A+B+C = D+E+F = 12; and the sum of each circle must be 16.A B C D E F 1 5 6 3 2 7 1 6 5 2 3 6 2 3 7 5 1 6 2 7 3 1 5 6 3 2 7 6 1 5 3 7 2 1 6 5 5 1 6 7 2 3 5 6 1 2 7 3 6 1 5 7 3 2 6 5 1 3 7 2 7 2 3 6 5 1 7 3 2 5 6 1 2. As the prime factors of the known points are {2, 3, 2*2, 3*3, 2*2*3}, x must be 2*3*3 = 18. As the square root of the product of 2*3*4*9*12*18 = 216, A*B*C = D*E*F = 216; the product of each circle is 1296 [36^{2}] A B C D E F 2 9 12 4 3 18 2 12 9 3 4 18 3 4 18 9 2 12 3 18 4 2 9 12 4 3 18 12 2 9 4 18 3 2 12 9 9 2 12 18 3 4 9 12 2 3 18 4 12 2 9 18 4 3 12 9 2 4 18 3 18 3 4 12 9 2 18 4 3 9 12 2 Quote Link to post Share on other sites

0 jasen 4 Posted January 26, 2016 Author Report Share Posted January 26, 2016 @DejMar 1st question 1 5 6 3 2 7 = 1 6 5 2 3 6 = 5 1 6 7 2 3 = 5 6 1 2 7 3 = 6 1 5 7 3 2 = 6 5 1 3 7 2 1st solution 2 3 7 5 1 6 = 2 7 3 1 5 6 = 3 2 7 6 1 5 = 3 7 2 1 6 5 = 7 2 3 6 5 1 = 7 3 2 5 6 1 2nd solution 2nd question 2 9 12 4 3 18 = 2 12 9 3 4 18 = 9 2 12 18 3 4 = 9 12 2 3 18 4 = 12 2 9 18 4 3 = 12 9 2 4 18 3 = 1st solution 3 4 18 9 2 12 = 3 18 4 2 9 12 = 4 3 18 12 2 9 = 4 18 3 2 12 9 = 18 3 4 12 9 2 = 18 4 3 9 12 2 = 2nd solution Just Rotation and reflection of the circle. Quote Link to post Share on other sites

0 DejMar 9 Posted January 26, 2016 Report Share Posted January 26, 2016 @jasen - Rotations and reflections may be considered the same for the visual layout, but not for the assignment of values to specific points. There are twelve different solutions. If one considers rotations and reflections to be pertinent, so one should one allow the eversion of this particular figure due to the placement of the points. In such case there is but one solution. Quote Link to post Share on other sites

0 jasen 4 Posted January 27, 2016 Author Report Share Posted January 27, 2016 Another "Easy" Puzzle Find The value of x and y, then put numbers 2,4,8,32,x,y to each point, so : I. The product of points in every circle are equal, (A*E*F*C = B*E*D*C = A*D*F*B) II. A*B*C = D*E*F There are 2 solutions (Reflection and rotation solutions are considered as 1 same solution) Quote Link to post Share on other sites

0 jasen 4 Posted January 28, 2016 Author Report Share Posted January 28, 2016 hint : all the numbers are power of 2. Quote Link to post Share on other sites

0 jasen 4 Posted January 29, 2016 Author Report Share Posted January 29, 2016 1st hint : all the numbers are 2 raised to the power of something. (2 = 2^{1}, 4 = 2^{2}, 8 = 2^{3, }32 = 2^{4}) 2nd hint : Use exponent laws Exponent Laws: x^{m} * x^{n} = x^{m+n} Quote Link to post Share on other sites

0 jasen 4 Posted February 1, 2016 Author Report Share Posted February 1, 2016 Solution : Spoiler I call it Another "Easy" Puzzle, because it have a strong connection with my first question (Easy Puzzle). use solution from my first answer : A = 1, B = 5, C = 6, D = 3, E = 2, F = 7 then write it like this :A = 2^{1} , B = 2^{5} , C = 2^{6} , D = 2^{3} , E = ^{}2^{2} , F = 2^{7} so (A,B,C,D,E,F) = (2,32,64,8,4,128) This works by the Exponent Laws: x^{m} * x^{n} = x^{m+n} Quote Link to post Share on other sites

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## jasen 4

1. Easy PuzzlePut numbers 1,2,3,5,6,7 to each point, so :

I. The sum of points in every circle are equal, (A+E+F+C = B+E+D+C = A+D+F+B)

II. A+B+C = D+E+F

There are 2 solutions.

2 .Harder PuzzleFind the value of x, then put number 2,3,4,9,12,x to each point, so :

I. The product of points in every circle are equal, (A*E*F*C = B*E*D*C = A*D*F*B)

II. A*B*C = D*E*F

There are 2 solutions.

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