A Pre-RMO question! -21

Let $\lfloor\frac{x}{5}\rfloor=k=\lfloor\frac{x}{7}\rfloor$ $k\leq \frac{x}{5}<k+1$ $5k\leq x<5k+5.......[1]$ $k\leq\frac{x}{7}<k+1$ $7k\leq x <7k+7.......[2]$ From $[1]$ and $[2]$ we get $7k<5k+5\Rightarrow 2k<5$ $and$ $5k<7k+7\Rightarrow -2k<7 \Rightarrow 2k>-7$ $\Rightarrow -7<2k<5$ As $x\geq0\Rightarrow k\geq0\Rightarrow 2k\in$ { $0,1,2,3,4$ } as $k\in Z^+\Rightarrow k\in$ { $0,1,2$ }

For $k=0$

From $[1]$ $0\leq x<5.......[A_0]$ From $[2]$ $0\leq x<7.......[B_0]$ From $[A_0],[B_0]$ $0\leq x<5.......[I]$ For $k=1$

From $[1]$ $5\leq x <10.......[A_1]$ From $[2]$ $7\leq x <14.......[B_1]$ From $[A_1],[B_1]$ $7\leq x <10.......[II]$ For $k=2$

From $[1]$ $10\leq x <15.......[A_2]$ From $[2]$ $14\leq x <21.......[B_2]$ From $A_2,B_2$ $14\leq x <15.......[III]$

From $[I],[II]$ and $[III]$ we get $x\in$ { $0,1,2,3,4,7,8,9,14$ }

As there are $9$ possible non-negative integer values of $x$ , $\therefore$ the answer is $9$

Let $f(x) = \left \lfloor \dfrac x5 \right \rfloor - \left \lfloor \dfrac x7 \right \rfloor$ . $f(x)$ has solutions within $\dfrac x5 - \dfrac x7 < |1|$ . For $x \le 0$ , we have $0 \le \dfrac x5 - \dfrac x7 < 1$ . It is obvious that $x=0$ is a solution. The upper bound of $x$ is given by $\dfrac x5 - \dfrac x7 < 1 \implies x < \dfrac {35}2$ and for integer $x$ , $x \le 14$ . Note that $x=14$ is also a solution. But not all integers from $0$ to $14$ are solutions. Those not solutions are when $f(x) = \left \lfloor \dfrac x5 \right \rfloor - \left \lfloor \dfrac x7 \right \rfloor = 1$ . They are $5$ and $6$ , when $\left \lfloor \dfrac x5 \right \rfloor = 1$ and $\left \lfloor \dfrac x7 \right \rfloor = 0$ , and $10$ , $11$ , $12$ , and $13$ , when $\left \lfloor \dfrac x5 \right \rfloor = 2$ and $\left \lfloor \dfrac x7 \right \rfloor = 1$ . Therefore the number of integer solutions is $15 - 6 = \boxed 9$ .