7. For the function $y=f(x)$, it is known that it is defined and continuous on the entire number line, odd, and periodic with a period of 5, and that $f(-1)=f(2)=-1$. What is the minimum number of roots that the equation $f(x)=0$ can have on the interval [1755; 2017]? Answer: 210.
Solution. Since the function $f$ is odd and defined at zero, we get $f(0)=-f(0)$ $\Rightarrow f(0)=0$. Due to the 5-periodicity, we then have $f(5)=f(0)=0$. Using the oddness again: $f(1)=-f(-1)=1$, and due to the 5-periodicity $f(3)=f(-2)=1, f(4)=$ $f(-1)=-1$. Thus, at points $1, 2, 3$, and $4$, the values of the function are $1, -1, 1$, and $-1$, respectively. Therefore, on each of the three intervals between these points, there is at least one zero of the function $f$. Hence, on the period $[0; 5)$, the function has at least 4 zeros (it is clear that this estimate is achievable: for example, one can take a piecewise-linear function, which will have exactly 4 zeros). On the interval $[1755; 2015)$, the period fits 52 times (on it, there are at least $52 \cdot 4=208$ zeros), plus the zero at the point 2015 and at least one on the interval (2016; 2017). In total, there are at least 210 zeros.