Friday Puzzler -- Pogo Stick

Re: Friday Puzzler -- Pogo Stick

Larry Barrett

Question - does jumping over a space where a stone used to be count as one of the 6 stones in the jump? I assume so, but thought I would ask.

100^3 = 1,000,000, so one way to figure this out is to create a list of the 100 cubes and start jumping 6 numbers at a time and see what happens.

If Alex didn't throw in the extra jump (7 instead of 6 after jumping over a spot equal to a cube) it is easy to see that you would drown when you hit the first spot that is a multiple of 6, which would be 216, (216=6^3). On your 35th jump you would be at 210 and the next one would land on the hole where the 216th stone was.

But in this case, with the extra jump after jumping over the first 5 cubes, you would be on the 210+5 = 215 spot and would then jump over the 216 hole.

Here is the sequence of stones you would land on, starting from the island: 6, 13, 20, 26, 33, 39, 45, 51, 57, 63, 69, 76, 82, ...

The stones in red are 7 greater than the previous number because you jumped over the 1, 8, 27, and 64 spots, the first 4 cubes.

Modulo 6 arithmetic tells you what the remainder is as you divide numbers by 6. When you do this for the first 6 cubes you get remainders of 1, 2, 3, 4, 5, 0. For example, divide 8 by 6 and the remainder is 2, divide 27 by 6 and the remainder is 3. To apply this to our problem, you can see that after you jump over the 8th stone (an empty spot because it is a cube), and before you reach the 27th spot, you are 2 stones ahead of some multiple of 6. For example, after 3 jumps you are on the 18+2 stone, and after the next jump you are on the 24+2 stone. If 3^3 was equal to 26 you would land in the hole and drown. But 3^3=27 so you keep on jumping.

From looking at your location just before you reach each of the the first 6 cubes (after the jump anyway) it looks like you always land 1 stone before the next cube, and then jump over it. So I think you make it to the far shore OK.

This is obviously difficult to explain, even if I am right. There must be an easier way to look at this problem.

Correct!

Alexy

Great observation that the successive cubes are 1,2,3,4,5,0 modulo 6.

To prove it, look at the difference between two successive cubes.

(n+1)^3 - n^3

= n^3 + 3n^2 + 3n + 1 - n^3

= 3(n)(n+1) + 1

Since either n or (n+1) is even, this says that the difference between two successive cubes (which is also the distance between the safe stones one before a cube) is one greater than a multiple of 6, i.e., is 1mod6.