Ten checkers numbered 1 – 10 are lined up as shown in the figure below. Your objective is to create five piles of two checkers each in only five moves. On each move a single checker must “jump” exactly two checkers (in either direction) and land on the next single checker. The two jumped checkers may be two single checkers side-by-side or a stacked pair.
Hints:
Try to use 10 pennies represent checker 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
examples: checker 1 was on top checkers 3 and so on.....
Have fun with it.
I think this will help you to solve it, thanks for interested in.
I will unlock this thread on the 1st of October (assuming I remember), and
if there are no correct solutions within 24 hours I will post one of the solutions.
Gratis clue: the gaps left by checkers that have jumped are not significant
and are to be treated as though they are not there.
RonL
I have reopened this early as I may not be around to do this
when I said I would.
This is a variant of the eight coins problem.
1. Start by jumping checker 7 over 8 and 9 onto 10.
Now we are left with an eight checker problem with checkers 1, 2, 3, 4,
5, 6, 8, 9
.
2. jump 4 over 5&6 onto 8
3. jump 6 over 5&3 onto 2
4. jump 3 over stack with 6&2 onto 1
5. jump 5 over stack with 4&8 onto 9.
(note the gaps between checkers left by jumps are ignored for subsequent
moves)
This generalises to any even number of checkers >=8. It is fairly easy
to show that there is no solution for 6 or fewer checker.
RonL
my solution is:
step 1: let checker 7 jumps over 8 & 9 onto 10
1 , 2, 3, 4, 5, 6, 8, 9, 7/10
step2: jump checker 5 over 3 & 4 onto 2
1, 5/2, 3, 4, 6, 8, 9, 7/10
step 3: jump 3 over 4 & 6 onto 8
1, 5/2, 4, 6, 3/8, 9, 7/10
step 4: jump 9 over 3/8 onto 6
1, 5/2, 4, 9/6, 3/8, 7/10
step 5: the last move will jump the checker 1 over 5/2 onto 4.
I hope this will be the same as yours. No arguments huh ? Let's others opinions. Thanks anyways Captainblack.