Over the weekend I picked up a 7 block math snake. There are seven wooden blocks that each turn, four are numbers 1-4, two are operators and one ‘=’ sign. The point of the puzzle is to make an equation on all four sides that work. The only catch is that the equation is read left to right so order of operations is not used. So 3+3x3=18 and not 12.
My snake looks like this, each row is done by pushing the snake away from me.
1 + 1 + 1 = 1
3 x 3 x 3 = 2
4 / 2 / 4 = 4
2 - 4 - 2 = 3
Anyone want to try their hand at solving this one? They say there’s only one solution. I’d like to see how other people figure something like this out.
I have a feeling brute force is required. First, determine every possible way to get 1 (e.g. 3 + 2 - 4, or 1 / 3 x 3), then see if the other three rows have correct solutions.
Unfortunately, I count 73 different ways to form an equation with “= 1” at the end, and may have missed some.
Well, brute force is kind of required, but some equations you know will be thrown out right away. Anything that would end in 0 or negative numbers, so 1-1x1 wouldn’t work. Most equations that would start in a fractions, 2/3, isn’t going to work out well either.
I don’t know how easy it would be to solve without it sitting in front of you. Just thought people around here might be interested in it.
Found it:
Answer
4-33=3
2+2/4=1
14-2=2
3/1+1=4
Process
[spoiler]I started with the =3 result rather than =1, on the grounds that it would have the smallest number of results using multiplication or division. I laid out the sixteen different combinations of operators and worked out how many equations worked for each pair - 55 in total.
Then I went down one step for each pair of operators and listed what the next side’s operators would be, and what they would have to equal (1). Then I looked at what numbers would be on that side and filled them in to see if they equaled 1. Each time they did (a total of 4 times before I found the answer) I repeated the process, looking at the next pair of operators and seeing if they equaled 2.[/spoiler]
I’m glad someone got it. I did it a bit differently:
I started working with the 4 on the left. My reasoning was that the majority of the + and x operations would be too big. I then went through all the possible combinations of 4+1+1, 4+1+2, etc. Though I saw right away that many of the formulas wouldn’t work right off the bat. I also wrote down part of the formula so I could cross it off, made it far easier for me to see what I had done and what wouldn’t work right away. My paper looked like this
4+1+
4+1-
4+1/
4+1x
4+2+
4+2-
4+2/
4+2x
I could see right away that all the 4+x+ couldn’t work, the same with 4+Zx couldn’t work. I think it took 10-15 minutes for me to finish it this way. There is at least one or two combinations that solve 3 of the 4 sides. Sucks to think you’ve solved it only to get to the fourth side and see, no, I haven’t. Now I’m kind of interested to see if I can solve the 9 cube version, though that might make me tear my hair out.