School-Safe Puzzle Games

Jumping Pennies

penny coin puzzle

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 :)

11 Comments to “Jumping Pennies”


  1. suineg | Guest

    I 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


  2. RK | Profile

    Hi there, Suineg- to clarify a bit: you have to pass over 2 pennies per jump-either 2 in a row, or 2 stacked


    thanks


  3. suineg | Guest

    ok, 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.


  4. Tommy | Guest

    It 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.


  5. Will | Guest

    Can you jump 1 penny twice?


  6. Will | Guest

    If 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


  7. suineg | Guest

    RK what about my second approach jumping just 2 coins? Is it good?


  8. suineg | Guest

    I 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 ;>;)


  9. suineg | Guest

    the middle solution is bad, only the symmetricals seem to work, sorry;..: (


  10. suineg | Guest

    yeah, 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


  11. RK | Profile

    Suineg- I think you got it with your second approaches, nice job


    Clever, Tommy :)


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