[(1/3)^6÷(1/3)^5]÷1/3
Answers
\huge \bold{ \fbox{\red{required \: answer}}}
The answer can be literally anything. I can see most of the answers says 3 but, that is only one of the solutions. It is the answer when questions defines the relationship as number of characters, but the question never says that. All it says is, find me a relation which satisfies few values, and then find the relationship value at a new point.
Let us start by examining the question. So here is the question in mathematical terms:
Given:
f(1)=3, f(2)=3, f(3)=5, f(4)=4 and f(5)=4
Find f(6).
So if we assume the function to be f(x) then we know its value at 5 points.
Let’s say f(6)=K, so now we know it’s value at 6 points.
Now, given the value of a function at N points, we can always find a (N-1) degree polynomial which will be that function.
So for this particular case, I will choose f(x) = [math]ax^5+bx^4+cx^3+dx^2+ex+g[/math]
So I can write:
[math]\begin{bmatrix}a\\b\\c\\d\\e\\g\end{bmatrix} = \begin{bmatrix}1&1&1&1&1&1\\2^5&2^4&2^3&2^2&2&1\\3^5&3^4&3^3&3^2&3&1\\4^5&4^4&4^3&4^2&4&1\\5^5&5^4&5^3&5^2&5&1\\6^5&6^4&6^3&6^2&6&1\end{bmatrix}^{-1} \begin{bmatrix}3\\3\\5\\4\\4\\k\end{bmatrix}[/math]
Now, you can put whatever value of K you want, and you will get corresponding set of coefficient vector which will satisfy the function values at the given points.
On a side note, I don’t consider this puzzle a good puzzle, since most of the people asking it expect only 3 as answer. They basically want you to limit the way you think, in a particular direction, which I think is not a very great idea.