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Prime Hopscotch (Posted on 2010-01-06) Difficulty: 4 of 5
Replace the numerals 1 through 8 with ever increasing prime numbers, always using the next lowest possible that is available1 to fulfill the criteria on the left. Then do the same for the right.

Present a series for the left, and one for the right.

7 + 8 = Square

6 = Prime

4 + 5 = Square


1 + 2 + 3 = Square


















7 + 8 = Cube

6 = Prime

4 + 5 = Cube


1 + 2 + 3 = Cube






If it was required that the Right set required "1" to be the next Prime following on after that used for the "8" in the Left set, what might the Right set read, if indeed it is possible?

1. Note, "always using the next lowest possible that is available" means that if it is next on the list it cannot be dismissed unless it is the last of a group of two or three and will not fulfill the criterion. Only then may you advance to the next.

See The Solution Submitted by brianjn    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution solution dependant upon interpretation | Comment 1 of 17
Under the interpretation that what is sought is a set of primes  where P1< P2 < P3 < P4 < P5 < P6 < P7 < P8, beginning with the lowest possible value of P1, where all criteria may be met; with the additional criteria being that for each series: P1+P2+P3, P4+P5 and P7+P8, each are perfect powers (squares for the left and cubes for the right), the following is the solution: 

Left (2, 3, 11, 13, 23, 29, 31, 131)
L1 + L2 + L3 is a Square; 2 + 3 + 11 = 16 = 42
L4 + L5 is a Square; 13 + 23 = 36 = 62
L7 + L8 is a Square; 31 + 131 = 144 = 122

Right (2, 3, 59, 61, 1667, 1669, 1693, 4139)
R1 + R2 + R2 is a Cube; 2 + 3 + 59 = 64 = 43
R4 + R5 is a Cube; 61 + 1667 = 1728 = 123
R
7 + R8 is a Cube; 1693 + 4139 = 5832 = 183

Edited on January 7, 2010, 8:26 am
  Posted by Dej Mar on 2010-01-06 14:52:27

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