creepercraft97T3

They could also be trinary functions, if you want the base to also change around.

The hyperoperations are basically a sequence that extends the 3 basic functions everyone is familiar with.

Addition is hyperoperation 1

Multiplication is hyperoperation 2

Exponentiation is hyperoperation 3

Tetration is hyperoperation 4

and it continues on forever (although hyperoperations greater than or equal to 4 have problems with their domains).

Let f(x) be the 'n'th hyperoperation. You can get from hyperoperation 'n' to hyperoperation 'n+1' by setting the 'x' of f(x) to a certain value and then iterating f(x) 'x' times. That takes a lot of work, but for going from hyperoperation 'n' to hyperoperation 'n-1', you just need to use the formula $f(1+{f}^{-1}(x))$, where $f^{-1}(x)$ is the functional inverse of $f(x)$. Using this, you can find out how to go from hyperoperation 'n' to hyperoperation 'n-2' or other values. Basically, you can go from hyperoperation 'n' to hyperoperation 'n-v', where 'v' is an integer.

Hyperoperation 'n-2' from hyperoperation 'n': $f(1+f^{-1}(1+f(-1+f^{-1}(b))))$

I then remembered something I did about a year ago. I set up this equation $f(a)=f^2(b)$ and solved for $f(b)$, and I got $f^{0.5}(a)=f(b)$. Then I did it for the 3rd iteration, then the 4th, and so on. In other words, I learned how to get $f^{\frac1n}(x)$ from $f^n(x)$. So I figured that I could apply this to hyperoperations.

Basically, if you set up this equation $g(1+g^{-1}(a))=f(1+f^{-1}(1+f(-1+f^{-1}(b))))$ ('n-1' and 'n-2' equations equal to each other) and manage to solve the right side for $f(1+{f}^{-1}(x))$ (the 'n-1' equation), the left side will, consequently, turn from an 'n-1' equation into an 'n-0.5' equation. This would give us the ability to define fractional hyperoperations (and not just for 'n-0.5'), and if a useful, usable pattern is found, can lead to the generalization of all hyperoperations.

I actually don't know the answer (or even how to go about solving it), so I asked the question to see if anyone else could figure it out. However, if it is solved, then the results could be groundbreaking.

Flip the x and the y, so it becomes $x=3y+12$

Then just solve for y

$x-12=3y+12-12$

$\frac{x-12}{3}=\frac{3y}{3}$

$\frac{x}3-4=y$

So the inverse is $\frac{x}3-4=y$

Well, we can eliminate a few answers right away. 103% is not possible, so it's not that. And 95 and 91 are clearly too low if you flip it 200 times (but who would have the patience to do that?). However, it's also pretty clear that it would round to 100% instead of 99%. The reason is because they might have truncated the percent instead of rounding it.

I actually proved 4 equations in 8th grade, and #4 helps out a lot with this.

#4: $1-(1-c)^t$ the probability of an event with $c$ chance occurring at least once after $t$ times.

For your problem, $t=200$ and $c=\frac12$. $1-(1-\frac12)^{200}=1-(\frac12)^{200}=1-\frac{1}{BIG}=\frac{BIG-1}{BIG}$. I substituted 'BIG' for the result of the power because the number itself is 61 decimal digits. If we transfer the fraction generated into a percent and truncate at 60 decimal digits, then this is the percent: $99.999999999999999999999999999999999999999999999999999999999937$%. If I round, it will without a doubt round to $100$%. But if I truncate the entire decimal, I get $99$% (if you are wondering, truncating a fraction is basically cutting off its decimal after a certain point. In this case, I cut off all of its decimal, and probably so did the textbook.).

37 and 73 are also part of a smaller sequence. The list contains prime numbers in which every single permutation (including deletion) is prime

SEQUENCE:

2-2

3-3

5-5

7-7

37-37,73,3,7

73-73,37,7,3

There are only these 6 numbers in the list.

Answer to your question (if the second number isn't double-digit): 10

Answer to your question (if the second number is double-digit): 12