How long did it take you to solve this "maths equation"

KMeriwetherD · November 14, 2016, 6:51pm

BWYF!!! yeah, I never would have gotten it. I’m rusty on sequences and never would have thought to go there. I would like to review them and see if I concur with your answer just for fun. Maybe for my birthday next month I will go somewhere tropical, lay on a beach, and learn sequences.

Well done (I’m assuming) BRAVO

brain_wash_your_face · November 14, 2016, 8:25pm

If it was something other than Fibonacci, I would have had a hard time. Fibonacci is ingrained in me always thought is was pretty cool:

Mobius_Strip · November 14, 2016, 8:44pm

I thought your “operating assumption” was diamond=1. You concluded that diamond=5 later, which calls into question the validity of some earlier steps (particularly when you were finding solutions to the equation a+4b+c=18).

Also, we end up f(5)=5 and f(8)=21, so we conclude that f(n)=n(th) Fibonacci number? Great job for picking up this relatively obscure pattern from just two examples, but without any additional information on constraints there are endless “solutions” to what f(n) might be.

But I guess like others have mentioned earlier, this is not a math problem but a pattern-finding problem where you are free to make up the rules and find any internally-consistent logical system that fits the symbols. So in that regard, you clearly solved it.

brain_wash_your_face · November 14, 2016, 9:05pm

No, I was just trying to steer people in the correct direction, and they had previously assumed diamond = 1. At that point, they were working on the circle, square, triangle part of the problem. a and c cannot both equal 1 because they would then not satisfy the other equations.

You are correct that there are other functions that could satisfy f(5)=5 and f(8)=21. Care to throw some out there?

Mobius_Strip · November 14, 2016, 9:50pm

how about f(n)=[(1 + sqrt(5))^n - (1 - sqrt(5))^n]/[sqrt(5)*2^n]

brain_wash_your_face · November 14, 2016, 10:19pm

Yep, that definitely satisfies the those two equations. I wouldn’t really want to have to calculate the last equation, though: f(f(n) would be the below. Yuk. I did think about that though, there are certainly infinite solutions. However, I’d be surprised if the below was the answer for which they are looking.

[(1 + sqrt(5))^[(1 + sqrt(5))^n - (1 - sqrt(5))^n]/[sqrt(5)*2^n] - (1 - sqrt(5))^[(1 + sqrt(5))^n - (1 - sqrt(5))^n]/[sqrt(5)*2^n]]/[sqrt(5)*2^[(1 + sqrt(5))^n - (1 - sqrt(5))^n]/[sqrt(5)*2^n]]

KMeriwetherD · November 24, 2016, 10:10pm

brain\_wash\_your\_face:

Since nobody has follwed up I’ll post the answer to the below problem. Spoilers:

_ Spoilers: _

Easy Functions:

Sequential shapes are addition. Example: square square = square+square

Enveloping (big) circle = squareroot(x)

Enveloping triangle = x^2

Using these functions on equations one, three, five and seven you will find that circle = 2, square = 3, triangle = 4.

Here is where it starts to get tricky. What is diamond? We can use equation 6 to find diamond:

(diamond)^2+(diamond+3)^2+3 = (diamond+4)^2+3^2+2

diamond^2+(diamond+3)^2 = (diamond+4)^2+8

diamond = either -3 or 5. Given all other numbers are positive, lets use 5 and move on.

So, diamond = 5. We have all of the numbers to solve this. There is one function remaining. Here is where it gets damn tricky. What is function diamond(x)? We now need to find equations 2 and 4, both of which use diamond(x).

Equation 2: diamond(5) = 5.

Equation 4: diamond(8) = 21.What could solve both of these?

Here is the hard part. It s a Fibonacci series:

diamond(5) = 1,1,2,3,_ 5 _

diamond(8) = 1,1,2,3,5,8,13,_ 21 _

so, diamond(x) = xth Fibonacci value.

So the _ answer _ is:

Fibonacci(Fibonacci(5+2) =

Fibonacci(1,1,2,3,5,8,_ 13 _)

Fibonacci(13) = (1,1,2,3,5,8,13,21,34,55,89,144,_ 233 _)

Final answer = 233

Finally had a chance to go over this step by step. It does indeed work out. Really cool! Thanks for enlightening us. PA… give us another one…