Gif 1

Anti solution: The equation $\left \lfloor x \right \rfloor + \left \lfloor 5x \right \rfloor + \left \lfloor 10x \right \rfloor + \left \lfloor 20x \right \rfloor$ essentially is the line $y=36x$ divided into small horizontal lines of length $l= \frac{1}{20}$ spaced apart. Hence, if there exists one real solution there are infinite real solutions and the answer would be infinity. It can't be infinity, so there are no solutions and the sum is 0. Aren't anti solutions fun?

---

Real solution: As stated before, the equation $\left \lfloor x \right \rfloor + \left \lfloor 5x \right \rfloor + \left \lfloor 10x \right \rfloor + \left \lfloor 20x \right \rfloor$ essentially is the line $y=36x$ divided into small horizontal lines of length $l =\frac{1}{20}$ . So, let's consider the spacings between these horizontal lines.

First, I will show that the function has a period of $36$ .

$\left \lfloor x \right \rfloor + \left \lfloor 5x \right \rfloor + \left \lfloor 10x \right \rfloor + \left \lfloor 20x \right \rfloor = a$

Then $x+1$ has the same fractional part and hence $\lfloor bx \rfloor$ increases by $b$ . $1+5+10+20=36$ , hence you can obtain $a+36$ using $x+1$ . For the unobtainable values, note that every $x$ corresponds to just one value. If some $a$ cannot be obtained, any placement of a strip at height $a$ would result in an $x$ having multiple values. Since we just showed that the existing strips are periodic, then it follows that $a+36$ has the same issue and cannot be obtained.

Hence we check that $-1$ cannot be obtained (because if a<0 then the floor of a is at most -1) and due to this $36K+35$ cannot be either.

It can easily be proved that [nx] can have maximum value n[x] + n-1 by using x = [x} + {x} Now, the given sum lies between 36[x] and 36[x] + 34 but it is equal to 36k + 35. As k is an integer, both limits cannot be satisfied simultaneously. Hence, no solution exists

Let $x=I+f$ where $I$ is the integral part of $x$ and $f$ is the fractional part and $0 \leq f <1$

Case1: $0\leq f<\frac{1}{20}$

$\lfloor x \rfloor + \lfloor 5x \rfloor + \lfloor 10x \rfloor + \lfloor 20x \rfloor = 36K+35\\$ $\lfloor I+f \rfloor + \lfloor 5I+5f \rfloor + \lfloor 10I+10f \rfloor + \lfloor 20I+20f \rfloor = 36K+35\\$ $I+\lfloor f \rfloor +5I+ \lfloor5f \rfloor + 10I+\lfloor10f \rfloor +20I+ \lfloor 20f \rfloor = 36K+35\\$

Because $0\leq f<\frac{1}{20}$ $\implies \lfloor f \rfloor = 0,\lfloor5f \rfloor = 0,\lfloor 10f \rfloor = 0,\lfloor 20f \rfloor = 0$

Now on putting we get $I+5I+10I+20I=36K+35\\$ $36I=36K+35\\$

This yields no solution as $I$ and $K$ are integers.

Case2: $\frac{1}{20} \leq f<\frac{1}{10}$

$\lfloor x \rfloor + \lfloor 5x \rfloor + \lfloor 10x \rfloor + \lfloor 20x \rfloor = 36K+35\\$ $\lfloor I+f \rfloor + \lfloor 5I+5f \rfloor + \lfloor 10I+10f \rfloor + \lfloor 20I+20f \rfloor = 36K+35\\$ $I+5I+10I+20I+1=36K+35\\$ $36I=36K+34\\$

which is not possible as $I$ and $K$ are integers.

Case3: $\frac{1}{10}\leq f<\frac{1}{5}$

$\lfloor x \rfloor + \lfloor 5x \rfloor + \lfloor 10x \rfloor + \lfloor 20x \rfloor = 36K+35\\$ $\lfloor I+f \rfloor + \lfloor 5I+5f \rfloor + \lfloor 10I+10f \rfloor + \lfloor 20I+20f \rfloor = 36K+35\\$ $I+5I+10I+20I+2=36K+35\\$ $36I+3=36K+35\\$

which is again not possible for integers $I$ and $K$ .

Case4: $\frac{1}{5} \leq f<1$

$\lfloor x \rfloor + \lfloor 5x \rfloor + \lfloor 10x \rfloor + \lfloor 20x \rfloor = 36K+35\\$ $\lfloor I+f \rfloor + \lfloor 5I+5f \rfloor + \lfloor 10I+10f \rfloor + \lfloor 20I+20f \rfloor = 36K+35\\$ $I+5I+10I+20I+7=36K+35\\$ $36I+7=36K+35\\$

This case also yields no solutions

So from all the cases we conclude there are no real solutions of $x$ to the given equation for constant integer $k$