Question

This one is tough. If Karl the kangaroo starts at position (0, 0, 0) on a grid. He can either make a short jump, which takes him one unit east, one unit south, and two units up; a medium jump, which takes him one unit east, three units north, and one unit down; or a long jump, which takes him two units east, one unit north, and three units up. Unfortunately, Karl has a cold, and occasionally sneezes. When he does, he sneezes so hard he moves one unit north and two units up! So, for example, if he takes one short jump, one long jump, and then sneezes, he ends up at (3, 1, 7). Karl wants to end up at (100, 100, 100). Given that Karl really does have a cold and has to sneeze at least once, what is the fewest number of jumps it will take him to reach his goal? Summary: short moves (1,-1,2), med moves (1,3,-1), Long moves (2,1,3), and sneeze (at least one) moves (0,1,2).

Answer 1

Summarised data, (feel free to correct if wrong):

((East/West),(North/South),(Up/Down))
Short jump: (1,-1,2)
Medium jump: (1,3,-1)
Long jump: (2,1,3)
Sneeze effect: (0,1,2)

Must sneeze at least once as well.

Answer 2

first step: apply the sneeze to the (100,100,100)

Answer 3

(100, 100, 100) -> (100, 100-1, 100-2) -> (100,99,98)

Answer 4

I'm wondering how this will work... I have a system equations... Four variables yet only three equations. I've tried representing it as a matrix but nothing comes to mind when solving it...
Since we are using linear combinations of jumps and sneezes... It would look like this.
\[a\left(\begin{matrix}1 \\ -1\\-2\end{matrix}\right)+b\left(\begin{matrix}1 \\3\\-1\end{matrix}\right)+c\left(\begin{matrix}2 \\ 1\\3\end{matrix}\right)+x\left(\begin{matrix}0 \\ 1\\2\end{matrix}\right)=\left(\begin{matrix}100 \\ 100\\100\end{matrix}\right)\]

Answer 5

thats good. you can create the matrix equation and find a solution maybe.

Answer 6

Note: a should be (1, -1, 2)

Answer 7

Yeah oops missed that! Does anyone know how to edit the matrix code or copy it for editing, I can't seem to access the Tex code for it.

Answer 8

According to this: http://www.math.odu.edu
When I put in that system of equations with the corrected (1,-1,2)
This problem has an infinite number of solutions which is fine I guess since we want the minimum jumps so... I'm really stumped here.
If you've heard of row reducing, it helps solve these types of problems, this is what happens:

a 0 0 -3d = 0
0 b 0 (-7/5)d = 20
0 0 c (11/5)d = 40

Answer 9

that doesnt use the sneeze though, so add in 5 sneezes, 15 of the short jump, 7 of the medium jump, and take away 11 of the long jump. That should do the trick.

Answer 10

sorry, x=d which is how many sneezes can happen.

Answer 11

@SmoothMath, anything you can spot here?

Answer 12

Your row reduce tells us a couple of things. It says that 20 medium jumps and 40 long jumps will get us to (100,100,100). But that doesnt use any sneezes. It also tells us that:

-3(short)+-7/5(medium)+11/5(long)=(sneeze)

But we can only deal with integers. so multiplying everything by five yields:

-15(short)-7(medium)+11(long)=5(sneeze).

So if we add 5 sneezes, we need to take away 11 long, but add in 15 short and 7 medium

Answer 13

it will till land us on (100,100,100)

Answer 14

What's the euqation that discribes the solution to the problem?

Answer 15

So 15 (shorts), 27(mediums), 29(longs), and 5 (sneezes) should do the trick. If we add more sneezes, the total jump count will go up, so this is the least number of jumps.

Answer 16

And, this includes sneezes

Answer 17

I don't think the Kangaroo can do negative long jumps... I think you understand this better than me so I'm not sure of my statement.

Answer 18

(100,99,98), and add in the sneeze equation

Answer 19

put it in a matrix or something, and someone tell me the equation that gives me the results, please

Answer 20

Ah wait, I think I get it! You're right joemath!

Answer 21

There is no way the kangaroo can only do one sneeze. the matrix solutions dont come out in integers in that case.

Answer 22

@joemath314159 I'm not assumign the kangaroo can only do one sneeze

Answer 23

I'm assuming the kangaroo does at least one sneeze

Answer 24

Heres the matrix.
http://www.wolframalpha.com/input/?i=rref%7B%7B1%2C1%2C2%2C0%2C100%7D%2C%7B-1%2C3%2C1%2C1%2C100%7D%2C%7B2%2C-1%2C3%2C2%2C100%7D%7D

Answer 25

btw, a sneeze is not a jump

Answer 26

*is a sneez a jump

Answer 27

No, it isn't a jump

Answer 28

I think this has been solved!

Answer 29

@joemath314159 , might've solved it, I never paid attention to matrices and I've never taken anything past trig (definitely nothing close to linear), so yea

Answer 30

its a linear algebra question at heart, which im very fond of. In linear algebra terms, the 3 jumps are linearly independent, and span all of R^3. The sneeze is linearly dependent, so it can be written as a linear combination of the 3 jumps. Since we are dealing with a 3 x 4 matrix, the system will automatically have infinitely many solutions, we just need to find a specific one. Very interesting indeed.

Answer 31

So what I understand from Joe:

d has to be a multiple of 5. At most, d=15 before c(the long jumps) becomes negative to solve for 40(which is impossible I think).

Therefore, putting in 5,10 and 15 sneezes tells us the total number of jumps required to arrive (100,100,100). We then look at which multiple of 5 gives us the least jumps.

Answer 32

This does seem a little complicated though, I'm not sure this is the approach the question intended XD!

Answer 33

Thank you all.