Minimum number of possible value of x Let $f: R\rightarrow R$ is a function stratifying $f(2-x)=f(2+x)$ and $f(20-x)=f(x)$ for all $x$ belonging to $R$ . If $f(0) =5$ , them minimum number of possible value of $x$ satisfying $f(x) =5$ for $x$ belonging $[0,170]$ 
 A: 
Define $g(x)=\frac{5}{2}(x-2)$ for $0\le x\le10$ and $\frac{5}{2}(18-x)$ for $10\le x\le18$, and extend it to all $x$ by making it periodic with period 16. 
It is easy to check that this satisfies the conditions $g(2-x)=g(2+x)$ and $g(x)=g(20-x)$.
It is clear that in any one period $g(x)$ has two values for which $g(x)=5$. For the interval $[2,18]$ they are $x=4$ and $x=16$. So each of the 10 complete periods in the range $[0,170]$, namely $[2,18],\dots,[146,162]$ contains two values of $x$. The values 0 and 164 are also in the range, so a total of 22.
Now suppose that $f(x)$ is any function satisfying $f(2-x)=f(2+x)$ and $f(x)=f(20-x)$. Then for any $x$ we have $f(16+x)=f(20-(16+x))=f(4-x)=f(2-(x-2))=f(2+x-2)=f(x)$, so $f(x)$ must be periodic with period 16.
Also we have $f(4)=f(2+2)=f(2-2)=f(0)=5$. So the 11 values $0,16,32,\dots,160$ and the 11 values $4,20,36,\dots,164$ are in the range $[0,170]$. Hence there are at least 22 values in the range where $f(x)=5$.
Hence the minimum is 22.
