Find the function which satisfies the given conditions Let $f :R \to R$  is a function satisfying $f(10-x) = f(x)$ and $f(2-x) = f(2+x)$

for all $x$ belonging to $R$.If $f(0) = 101$, then find the minimum possible no. 

of values of $x$ satisfying $f(x) = 101$ for $x$ belonging to $[0,30]$.

As $f(2-x) = f(2+x)$, hence $f(x)$ is symmetric about the line $x=2$ but I am not able to understand how to use other information. Could someone give me slight hint?
Hint (as requested): What can you say about  $f(5)$, using the first condition? 

What about $f(4)$? What's the second condition tell you about $f(0)$? What's the first condition tell you about $f(10)$? Do you notice a pattern here? 

Your question title is a bad one, because you cannot find "the function" satisfying these conditions -- there may be many many such functions.
By condition $f(2-x)=f(2+x)$ we can say $f$ is periodic with period $4$ and with condition $f(10-x)=f(x)$ means $f$ is periodic with period $10$ so as $f$ is the function satisfying both the properties so its period should divide $4$ and $10$ so period of $f$ will be either $1$, $2$ or any number. But as we want the minimum so period should be the maximum of all these so the period should be $2$ which implies it has $16$ values of in $[0,30]$ satisfying given conditions.