Variant 132. $\frac{1}{6}$. Variant 133. $\frac{5}{6}$. Variant 134. $\frac{2}{9}$.

Grading criteria: 20 points - correct (not necessarily the same as above) solution and correct answer; $\mathbf{1 5}$ points - an arithmetic error was made in the final calculation, provided the solution is otherwise correct; 5 points - the expression for efficiency is written, the relationship between the specified powers is noted, but the answer is not obtained.

Problem 5 (variant 131).

The entire terrain is divided into squares, marked by two integer indices $M$ and $N$ such that, for example, the point with coordinates $x=12.25, y=20.9$ is in the square numbered

$[12 ; 20]$, and the point with coordinates $x=-12.34, y=0.1239$ is in the square numbered $[-13 ; 0]$, and so on. A mysterious object moves in the Oxy plane along the trajectory $y=\left(\left(\left(x^{5}-2013\right)^{5}-2013\right)^{5}-2013\right)^{5}$, while the radar beam on the terrain is directed along the line $y=x+2013$. Indicate the numbers of all squares in which the radar will register the appearance of the mysterious object.
Solution. 1) The equation $\left(\left(\left(x^{5}-2013\right)^{5}-2013\right)^{5}-2013\right)^{5}-2013=x$ is equivalent to the equation $x^{5}-2013=x$.

Possible proof: a) If $x$ is a root of equation (2), then

$\left(\left(\left(x^{5}-2013\right)^{5}-2013\right)^{5}-2013\right)^{5}-2013=\left(\left(x^{5}-2013\right)^{5}-2013\right)^{5}-2013=\left(x^{5}-2013\right)^{5}-2013$ $=x^{5}-2013=x$, i.e., $x$ is a root of equation (1). b) If $x$ is a root of equation (1), then it must also be a root of equation (2). Suppose the opposite. Let $x^{5}-2013>x$. Then, from the fact that the function $g(x)=x^{5}-2013$ is monotonically increasing, it follows: $\left(\left(\left(x^{5}-2013\right)^{5}-2013\right)^{5}-2013\right)^{5}-2013>\left(\left(x^{5}-2013\right)^{5}-2013\right)^{5}-2013>\left(x^{5}-2013\right)^{5}-2013>$ $x^{5}-2013>x$, i.e., the left side of equation (1) is greater than the right side. Similarly, there are no solutions if $x^{5}-20130$, so due to continuity, there will be roots (one root, but this is not important here) between points 4 and 5.
3) The equation $x^{5}-2013=x$ has no roots outside the interval $(4 ; 5)$. We will prove this.

Since $f(x)=x\left(x^{4}-1\right)-2013$, there are no roots when $x \leqslant-1$ and when $x \in(1 ; 4)-$ in these cases, it is obvious that $f(x)5$, because then $x\left(x^{4}-1\right)-2013>5\left(5^{4}-1\right)-2013>0$.

The absence of roots on $(4 ; 5)$ can also be proven using the derivative: $f^{\prime}(x)=5 x^{4}-1-$ the function is increasing everywhere except for $x^{4} \leqslant \frac{1}{5}$ (some neighborhood of zero). Then, this neighborhood needs to be carefully "circumnavigated." For example, there are no roots on $x \in[-1 ; 1]$ (see above), no roots when $x \leqslant-1$ (see above), and there will be exactly one root when $x>1$.

4) Therefore: $x \in(4 ; 5)$. Hence, $y=x+2013 \in(2017 ; 2018)$.

Answer: $[4 ; 2017]$.

## Answers of other variants: