2. Zadatak

Prvo rješenje.

Uočimo da je lijeva strana početne jednakosti cijeli broj pa to mora biti i desna.

Stoga postoji $k\in\mathbb{Z}$ takav da je

$$\frac{5x-4}{3}=k.$$

Tada je

$$x=\frac{3k+4}{5}.$$

Uvrštavanjem u početnu jednadžbu nakon sređivanja dobivamo:

$$\left\lfloor\frac{2k+1}{5}\right\rfloor+\left\lfloor\frac{4k+7}{10}\right\rfloor=k.$$

Vrijedi $k=5a+b$, za $a\in\mathbb{Z}$, $b\in\{0,1,2,3,4\}$:

$$\begin{aligned} \left\lfloor\frac{2(5a+b)+1}{5}\right\rfloor+\left\lfloor\frac{4(5a+b)+7}{10}\right\rfloor&=5a+b\\ \left\lfloor\frac{10a+2b+1}{5}\right\rfloor+\left\lfloor\frac{20a+4b+7}{10}\right\rfloor&=5a+b\\ 2a+\left\lfloor\frac{2b+1}{5}\right\rfloor+2a+\left\lfloor\frac{4b+7}{10}\right\rfloor&=5a+b\\ \left\lfloor\frac{2b+1}{5}\right\rfloor+\left\lfloor\frac{4b+7}{10}\right\rfloor&=a+b. \end{aligned}$$

Razlikujemo 5 slučajeva:

1. $b=0$

   $$a+0=\left\lfloor\frac{2b+1}{5}\right\rfloor+\left\lfloor\frac{4b+7}{10}\right\rfloor=\left\lfloor\frac{2\cdot0+1}{5}\right\rfloor+\left\lfloor\frac{4\cdot0+7}{10}\right\rfloor=\left\lfloor\frac15\right\rfloor+\left\lfloor\frac7{10}\right\rfloor=0+0=0.$$

   U ovom slučaju je $a=0$, $b=0$, $k=0$, $x=\frac45$.

2. $b=1$

   $$a+1=\left\lfloor\frac{2b+1}{5}\right\rfloor+\left\lfloor\frac{4b+7}{10}\right\rfloor=\left\lfloor\frac{2\cdot1+1}{5}\right\rfloor+\left\lfloor\frac{4\cdot1+7}{10}\right\rfloor=\left\lfloor\frac35\right\rfloor+\left\lfloor\frac{11}{10}\right\rfloor=0+1=1.$$

   U ovom slučaju je $a=0$, $b=1$, $k=1$, $x=\frac75$.

3. $b=2$

   $$a+2=\left\lfloor\frac{2b+1}{5}\right\rfloor+\left\lfloor\frac{4b+7}{10}\right\rfloor=\left\lfloor\frac{2\cdot2+1}{5}\right\rfloor+\left\lfloor\frac{4\cdot2+7}{10}\right\rfloor=\left\lfloor\frac55\right\rfloor+\left\lfloor\frac{15}{10}\right\rfloor=1+1=2.$$

   U ovom slučaju je $a=0$, $b=2$, $k=2$, $x=2$.

4. $b=3$

   $$a+3=\left\lfloor\frac{2b+1}{5}\right\rfloor+\left\lfloor\frac{4b+7}{10}\right\rfloor=\left\lfloor\frac{2\cdot3+1}{5}\right\rfloor+\left\lfloor\frac{4\cdot3+7}{10}\right\rfloor=\left\lfloor\frac75\right\rfloor+\left\lfloor\frac{19}{10}\right\rfloor=1+1=2.$$

   U ovom slučaju je $a=-1$, $b=3$, $k=-2$, $x=-\frac25$.

5. $b=4$

   $$a+4=\left\lfloor\frac{2b+1}{5}\right\rfloor+\left\lfloor\frac{4b+7}{10}\right\rfloor=\left\lfloor\frac{2\cdot4+1}{5}\right\rfloor+\left\lfloor\frac{4\cdot4+7}{10}\right\rfloor=\left\lfloor\frac95\right\rfloor+\left\lfloor\frac{23}{10}\right\rfloor=1+2=3.$$

   U ovom slučaju je $a=-1$, $b=4$, $k=-1$, $x=\frac15$.

Dakle, rješenje je

$$x\in\left\{-\frac25,\frac15,\frac45,\frac75,2\right\}.$$

Drugo rješenje.

Za sve realne brojeve $r$ vrijedi $r-1<\lfloor r\rfloor\le r$ pa vrijedi:

$$\frac{2x-1}{3}-1<\left\lfloor\frac{2x-1}{3}\right\rfloor\le\frac{2x-1}{3}$$ $$\frac{4x+1}{6}-1<\left\lfloor\frac{4x+1}{6}\right\rfloor\le\frac{4x+1}{6}.$$

Zbrajanjem tih nejednakosti dobivamo

$$\frac{2x-1}{3}-1+\frac{4x+1}{6}-1<\frac{5x-4}{3}\le\frac{2x-1}{3}+\frac{4x+1}{6}.$$

(Uočimo da ova nejednakost nije ekvivalentna polaznoj jednadžbi!)

Sređivanjem dobivamo

$$-\frac52<x\le\frac72. \tag{*}$$

Lijeva strana početne jednakosti je cijeli broj pa to mora biti i desna. Stoga postoji $k\in\mathbb{Z}$ takav da je $\frac{5x-4}{3}=k$, tj. $x=\frac{3k+4}{5}$.

Zbog nejednakosti $(*)$ jedini kandidati za rješenje su

$$-\frac{11}{5},\ -\frac85,\ -1,\ -\frac25,\ \frac15,\ \frac45,\ \frac75,\ 2,\ \frac{13}{5},\ \frac{16}{5}.$$

Uvrštavanjem u početnu jednadžbu, vidimo da jednadžbu zadovoljavaju samo

$$x\in\left\{-\frac25,\frac15,\frac45,\frac75,2\right\}.$$