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    • A follower on Twitter solved this recreational maths puzzler

      and then asked this intriguing question

      How could the problem be changed to have multiple sets of correct values?


    • That's a very open question. Depending on what exactly we are allowed to change, I see a number of ways to do so.

      Can we remove some of the equations and change the rest? If so, I notice that Y^(X-X) always equals one, independent of the values for X and Y.

      Can we change what the symbols mean? That feels a bit like cheating, though...

      Can we move from "regular" to modulo arithmetics? It's been a while since I've last done anything with that, but I feel as if that would open up the problem quite a bit. :)

    • ”Mods” or modulo or modular arithmetic is a very cool branch of mathematics.

      Typically, mods are off-limits in recreational mathematics. There’s no explicit rule against it, however, problems tend to provide greater productive struggle if mods are not part of the solution.

    • The most uninteresting part of recreational maths is the answer. The most interesting part of recreational maths is the answer to the questions

      how did you come up with your solution?

      what were you thinking when you tried that approach?

      what other approaches did you consider or discard?

      And yet, when I share my recreational maths creations online, the only responses I get are the answers. The best part of grad school for me was the times I would spend in productive struggle with other mathematicians, trying to solve a challenging problem and sharing our thinking and strategies. It was those experiences that evolved my problem solving skills, allowing me to confidently profess that I am a mathematician. I am therefore going to share my thinking in how I constructed this maths puzzle, how to solve it, and how I would attempt to change it to having multiple solutions.

      Full disclosure: I have no idea of how to change things at this point, but playing with maths is part of the process.

      How I came up with this maths puzzler

      Make it look harder than it really is. The recreational maths puzzlers I create can typically be solved with a basic knowledge of Algebra 1. If you tried to solve the above problem as a system of equations, you could subtract out πŸ¦„^πŸ¦„ from the first and third equations and be left with

      πŸ’”^πŸ’” βž– πŸ’”^πŸ¦„ = 18

      Okay, now what?

      Or you can look at the first equation and use the time honored strategy of guessing numbers and adjust based on the results.

      (πŸ¦„^πŸ¦„) βž• (πŸ’”^πŸ¦„) = 13

      Okay, a good guess would be numbers less than 13.

      How about πŸ¦„=1 and πŸ’”=12?

      1 + 12 = 13 βœ…

      Now plug those values into the second or third equation. Do they work?

      NO ❌

      Okay, what have we learned that’s useful? We know that the exponent has to be higher than 1. So let’s try πŸ¦„=2 and solve for πŸ’”.

      (πŸ¦„^πŸ¦„) βž• (πŸ’”^πŸ¦„) = 13

      4 + πŸ’”^2 = 13

      πŸ’”^2 = 9

      πŸ’” = 3 or -3

      Plug these values in for the second or third equations and you find that

      πŸ¦„=2 and πŸ’”=3 is the only combination that works for all three equations.

      I try to make my puzzles solveable without the use of a calculator. Even though it looks like an intimidating problem at first glance, with the right approach it could be solved with mental math.

      How do you change the puzzle to have multiple solutions?

      It turns out it’s a lot harder to do than I realized, @Factotum.

      Here’s some of the approaches I tried:

      Attempt#1. Set πŸ¦„=0. Unfortunately,

      πŸ¦„^πŸ¦„ = 0^0 = ❌

      Attempt#2. Add an additional term +πŸ¦„ to the second equation.

      Result: ❌