## Jumping Pennies

Here’s a fun little puzzle, but you need a couple of coins.

Place 10 pennies in a row upon a table. Then taking up any one of the series place it upon some other, with this rule: that you pass over just 2 cents. Repeat this until there is no single penny left!

Can you solve in 5 steps?

(for our readers who don’t use US currency, you can really play with any coin. When you make the jump, you can only pass over 2x that coin’s value. Hope that makes sense

suineg| Guest November 22nd, 2007 - 3:01 pmI have a question:

1) the goal of the game is to not have a single penny alone and can pass 1 or 2 pennys maximun?

if the answer is yes, assume every “1″ is a penny and “2″ two pennys stacked, i think here is in 5 steps:

1..1..1..1..1..1..1..1..1..1 (note all reference below are over the penny postion in this initial state)

1)2..1..1..1..1..1..1..1..1–>put the first over the second

2)2..2..1..1..1..1..1..1–>put third over fourth

3)2..2..2..1..1..1..1–>put the fifth over the sixth

4)2..2..2..2..1..1–>put the seventh over the eight

5)2..2..2..2..2-> put the ninth over the tenth

RK| Profile November 22nd, 2007 - 8:49 pmHi there, Suineg- to clarify a bit: you have to pass over 2 pennies per jump-either 2 in a row, or 2 stacked

thanks

suineg| Guest November 23rd, 2007 - 6:47 amok, let’s see how to do that in 5 steps:

1…1…1…1…1…1…1…1…1…1

note: position of the “1″ in the order from above

1)1…1…1…1…1…1…1…1…2–>position 7+position 10

2)1…1…1…1…1…2…1…2–>position 4+position 8

3)1…2…1…1…2…1…2–>position 5+ position 2

4)2…2…1…2…1…2–>position 3 + position 1

5)2…2…2…2….2–>position 6 + position 9

the secret is to group in piles of 2, because is the only way of making half of movements of the number of elements in the list, so 10/2=5, it was harder than it look anyway, but cool, i think i got it.

Tommy| Guest November 25th, 2007 - 12:30 pmIt is quite easy if you work backwards. I hope that’s not considered cheating.

Here’s one solution:

1111111111

111111112

11111212

1211212

121222

22222

There are many other “solutions”, but they are all quite similar to this one.

Will| Guest November 26th, 2007 - 1:49 amCan you jump 1 penny twice?

Will| Guest November 26th, 2007 - 2:01 amIf you can jump 1 penny twice: (move 5)

_)-1..2..3..4..5..6..7..8..9..10

0)-1..1..1..1..1..1..1..1..1..1=> 1 =single 2 =double

1)-1.._..1..1..2..1..1..1..1..1=> 2 Jumps to 5

2)-1.._..1..1..2.._..1..1..2..1=> 6 jumps to 9

3)-1.._..1.._..2.._..2..1..2..1=> 4 jumps to 7

4)-1.._..1.._..2.._..2.._..2..2=> 8 jumps to 10

5)-_.._..2.._..2.._..2.._..2..2=> 1 jumps 3 twice, lands on 3

suineg| Guest November 26th, 2007 - 9:38 pmRK what about my second approach jumping just 2 coins? Is it good?

suineg| Guest November 26th, 2007 - 10:04 pmI think there are 3 possible solutions in 5 steps:

there is a point where you had to have 2 “1..2..1″ in any order, the only possibilities are:

1)1..2..1..2..1..2..1—> starts making a “2″ in the middle

2)2..1..2..1..1..2..1—> starts making a “2″ in the left extreme

3)1..2..1..1..2..1..2—> starts making a “2″ in the right extreme

2) and 3) are symetrical solutions, 1) is the inside out solution ;>

suineg| Guest November 26th, 2007 - 10:12 pmthe middle solution is bad, only the symmetricals seem to work, sorry;..: (

suineg| Guest November 26th, 2007 - 10:22 pmyeah, the middle version does not work, because it seems to be imposiible to arriving at that configuration following the rules: so i see only two solutions that are symmetrical

RK| Profile November 26th, 2007 - 11:32 pmSuineg- I think you got it with your second approaches, nice job

Clever, Tommy