## Solution.

Consider the graphs of the functions $f(x)=\left|\sin \left(\frac{1}{3} \pi x\right)\right|$ and $g(x)=\frac{1}{a} x$.

The period of the function $\sin \left(\frac{1}{3} \pi x\right)$ is $\frac{2 \pi}{\frac{1}{3} \pi}=6$, so the period of the function $f$ is $\frac{6}{2}=3$.

1 point

Notice that all solutions to the equation are non-negative because the graphs do not intersect to the left of the $y$-axis.

![](https://cdn.mathpix.com/cropped/2024_05_30_bcbf631b6726be7fa637g-23.jpg?height=534&width=1274&top_left_y=935&top_left_x=311)

On each of the intervals $[0,3\rangle,[3,6\rangle,[6,9\rangle, \ldots,[3(k-1), 3 k\rangle$, the graphs of the functions $f$ and $g$ intersect at two points. This means that for 600 solutions, 300 such intervals are needed, each containing two solutions.

The last interval in which the graphs can intersect is $[3 \cdot 299,3 \cdot 300\rangle=[897,900\rangle$, and for any $x>900$, the function $g$ must have a value greater than the function $f$. (or see the image)

Consider the graphs of the functions $f$ and $g$ for $x \geqslant 900$.

![](https://cdn.mathpix.com/cropped/2024_05_30_bcbf631b6726be7fa637g-23.jpg?height=486&width=965&top_left_y=2007&top_left_x=471)

Notice that the function $f$ on the interval [900,903] has a maximum value of 1 at $x=901.5$ because $f(901.5)=\left|\sin \frac{901.5}{2} \pi\right|=\left|\sin \frac{\pi}{2}\right|=1$.

The function $g(x)=\frac{1}{a} x$ has a value of 1 at $x=a$.

Let $a=902$, i.e., $g(x)=\frac{1}{902} x$. Then $g(901.5)=\frac{901.5}{902} < 1$ for all $x>901$, and since $f(x) \leqslant 1$, the graphs of $f$ and $g$ do not intersect for $x>901$. We will show that they do not intersect for $x \in [900,901]$ either.

Calculate the values of the functions $f$ and $g$ for $x=900$ and $x=901$.

$$
\begin{gathered}
g(900)=\frac{900}{901} \approx 0.9, \quad g(901)=1 \\
f(900)=\left|\sin \frac{900 \pi}{3}\right|=0, \quad f(901)=\left|\sin \frac{901 \pi}{3}\right|=\frac{\sqrt{3}}{2} \approx 0.866
\end{gathered}
$$

Clearly, $0.9 < g(x) \leqslant 1$ and $0 \leqslant f(x) \leqslant \frac{\sqrt{3}}{2}$ for all $x \in [900,901]$, so $f(x) < g(x)$ on this interval, and their graphs do not intersect.

We conclude that the largest natural number $a$ with the desired property is $a=901$.

Note: If a student proceeds to the conclusion that the last interval in which the graphs intersect (with justification), but then simply writes that $a=901$ without any justification or calculation, they receive a maximum of 5 points.

If a student has done all of the above and calculated the values of the functions $f$ and $g$ for at least two values of $x \in \{900,901,902,903\}$ (it is clear that they are verifying the inequality $f(x) < g(x)$ for $x \geqslant 900$), but still does not have a valid and complete justification, they receive a maximum of 8 points.

If a student shows that for $a=901$, $f(x) < g(x)$ for $x \leqslant 900$, but does not show that this is the largest such, i.e., that $a=902$ is not a solution, they lose 1 point.

## ZUPANIJSKO NATJECANJE IZ MATEMATIKE

## 4th grade - high school - B variant

## February 28, 2019.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.
## Task B-3.7.

Determine the largest natural number $a$ for which the equation

$$
\left|\sin \left(\frac{1}{3} \pi x\right)\right|=\frac{1}{a} x
$$

has exactly 600 different solutions.