# How do you solve $[x]+[2x]+[3x]=4x$ on $\Bbb R$?

Find the arithmetic average of all solutions $$x\in\Bbb R$$ of the equation $$[x]+[2x]+[3x]=4x,$$ where $$[x]$$ denotes the integer part of $$x$$ (e.g. $$[2.5]=2$$, $$[-2.5]=-3$$).

I tried solving this problem by looking at $$\{x\}$$ and writing for example $$[2x]$$ as $$2[x]$$ when $$\{x\}<1/2$$ and $$2[x]+1$$ when $$\{x\}\ge1/2$$. This lead to a lot of cases and after half an hour I literally couldn't go on any longer. I was thinking maybe I can somehow find the arithmetic average without actually knowing the solutions, but couldn't find any way to do that. Any help would be appreciated. :)

Thanks!

• Drawing the graph might give you some ideas of how to attack the problem: wolframalpha.com/input/… Commented Jul 21, 2019 at 18:12
• You can see that $6x-3< [x]+[2x]+[3x]\leq 6x$, and that $4x$ must be an integer, and go from there. It still is checking cases, but from experience, it's the most straightforward and optimal way Commented Jul 21, 2019 at 18:14

As @Jakobian suggests we have that $$6x-3\lt \lfloor x\rfloor+\lfloor 2x\rfloor+\lfloor 3x\rfloor\le6x$$ Hence any solutions would require that $$6x-3\lt4x\le6x$$ $$2x-3\lt0\le2x$$ $$x-\frac32\lt0\le x$$ $$0\le x\lt\frac32$$ $$0\le 4x\lt6$$ Also, the only valid solutions $$x$$ are such that $$4x=\lfloor x\rfloor+\lfloor 2x\rfloor+\lfloor 3x\rfloor\in\mathbb{Z}$$. Using the above range this means that the only possible solutions are if $$4x=0,1,2,3,4,5$$ or $$x=0,\frac14,\frac12,\frac34,1,\frac54$$ respectively. We can test each case separately giving the only solutions $$x=0,\frac12,\frac34$$ Hence the arithmetic average of the solutions is $$\overline{x}=\frac{0+\frac12+\frac34}3=\frac5{12}$$

• This makes sense. I got to 0≤x<3/2 at some point but didn't realize that 4x was an integer. Thanks for the help! Commented Jul 21, 2019 at 18:36
• Done. Thanks again! Commented Jul 21, 2019 at 19:10

Note that $$4x \in \mathbb Z$$ and hence $$x= k +\frac{r}{4}$$ for some $$k \in \mathbb Z$$ and $$r \in \{ 0,1,2,3\}$$.

Then, the equation becomes $$4k+r= k+[2k+\frac{r}{2}]+[3k+\frac{3r}{4}]=6k+[\frac{r}{2}]+[\frac{3r}{4}]$$ which is equivalent to $$r=2k+[\frac{r}{2}]+[\frac{3r}{4}].$$

Now, just solve this for each $$r \in \{ 0,1,2,3\}$$:

• if $$r=0$$ then $$0=2k \Rightarrow x=0$$
• if $$r=1$$ then $$1=2k \Rightarrow \mbox{ no solution}$$
• if $$r=2$$ then $$2=2k+1+1 \Rightarrow x=\frac{1}{2}$$
• if $$r=3$$ then $$3=2k+1+2 \Rightarrow x=\frac{3}{4}$$

$$[x] + [2x] + [3x] = 4x$$ means $$4x$$ is an integer and we have four cases:

$$x = m$$.

$$x = m + \frac 14$$.

$$x = m + \frac 12$$

$$x = m + \frac 34$$

Where $$m= [x]$$.

If $$x = m$$ then $$2x = 2m$$ and $$3x =3m$$ and

$$[m] + [2m] +[3m] = m + 2m + 3m = 6m =4m$$. So $$m=x= 0$$.

If $$x = m + \frac 14$$ then

$$x = m + \frac 14$$ and $$2x = 2m + \frac 12$$ and $$3x = 3m + \frac 34$$.

So $$[m+\frac 14] + [2m + \frac 12] + [3m + \frac 34] = m + 2m + 3m = 6m = 4(m + \frac 14) = 4m + 1$$ so $$2m = 1$$ and $$m = \frac 12$$. No solution.

if $$x = m + \frac 12$$ then

$$x =m+\frac 12$$ and $$2x = 2m + 1$$ nad $$3x = 3m + \frac 32$$ so

$$[m] + [2m + 1] + [3m + \frac 32] = m + 2m + 1 + 3m + 1 = 6m + 2 = 4(m+\frac 12) = 4m + 2$$.

So $$6m = 4m$$ so $$m =0$$ and $$x = m + \frac 12 = \frac 12$$.

If $$x = m + \frac 34$$ then

$$x = m + \frac 34$$ and $$2x = 2m + \frac 32$$ and $$3x = 3m + \frac 94$$.

So $$[m + \frac 34] + [2m + \frac 32] + [3m + \frac 94] = m + 2m + 1 + 3m + 2 = 6m + 3 = 4(m +\frac 34) = 4m + 3$$.

So again $$m = 0$$ and $$x = m +\frac 34 = \frac 34$$.

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Note: there is a unique $$\{x\}; 0 \le \{x\} < 1$$ and $$x = [x]+\{x\}$$.

If $$\{x\} < \frac 13$$ then $$2[x] < 2x=2[x]+2\{x\} < 2[x] + 1$$ so $$[2x] = 2[x]$$.

And $$3[x] < 3x = 3[x]+3\{x\} <3[x]+1$$ so $$[3x]=3[x]$$.

So $$[x]+[2x]+ [3x] = 6[x] = 4x = 4[x] + 4\{x\}$$ means

$$[x] =\frac 23 \{x\}$$. But $$0 \le \frac 23\{x\} < \frac 23$$ and $$[x]$$ is an integer so $$[x] =\frac 23\{x\} = 0$$ and $$x = [x]+\{x\} = 0$$.

Solution 1: $$x = 0$$ and $$[0]+[0]+[0] = 4*0$$.

If $$\frac 13 \{x\} < \frac 12$$ then $$2[x] < 2x=2[x]+2\{x\} < 2[x] + 1$$ so $$[2x] = 2[x]$$.

$$[x] + \frac 13 \le [x]+\{x\} =x < [x] + \frac 12$$

$$3[x]+1 \le 3x < 3[x] + 1\frac 12$$ so $$[3x] = 3[x]+1$$.

So $$[x]+[2x]+ [3x] = 6[x]+1 = 4x = 4[x] + 4\{x\}$$ means

$$[x] = 2\{x\} - \frac 12$$.

But $$-\frac 12 \le 2\{x\}-\frac 12 = [x] < 12$$ so $$[x] =0$$. And $$\{x\} = \frac 14$$. But $$\{x\} \ge \frac 13$$ so no solution.

If $$\frac 12 \le \{x\}< \frac 23$$ then

$$[x] + \frac 12 \le [x] + \{x\} = x < [x]+\frac 23$$ and $$2[x]+1 \le 2x < 2[x] + 1\frac 13$$ so $$[2x]= 2[x]1$$.

$$[x]+\frac 13 < [x]+\{x\} = x < [x] + \frac 23$$ so $$3[x]+1 < 3x < 3[x]+2$$ so \$[3x]=3[x]+14.

So $$[x]+[2x]+[3x] = 6[x] + 2 = 4x = 4[x] + 4\{x\}$$ means

$$[x] = 2\{x\} - 1$$. Now $$0 =2*\frac 12 - 1\le 2\{x\}-1 = [x] < 2*\frac 23 -1 = \frac 13$$ so $$[x] = 0$$. and $$\{x\} = \frac 12$$. So $$x = [x] + \{x\} = \frac 12$$.

Solution 2: $$x = \frac 12$$ and $$[\frac 12] + [1] + [1\frac 12] = 0 +1 + 1=2 = 4*\frac 12$$.

Finally if $$\frac 23 \le \{x\} < 1$$ then

$$[x]+\frac 12 < [x]+\{x\} = x < [x] + 1$$ so $$2[x] + 1 < 2x < 2[x]+2$$ so $$[2x] = 2[x]+1$$.

$$[x] + \frac 23 \le [x]+\{x\} =x < [x] + 1$$ so $$3[x]+1 \le 3x < 3x + 1$$ so $$[3x] = 3[x] + 2$$.

So $$[x]+[2x] + [3x] = 6[x] + 3 = 4x = 4[x] +4\{x\}$$ so

$$[x] = 2\{x\} -\frac 32$$.

So $$-\frac 16 = 2*\frac 23 - \frac 32 < 2\{x\} -\frac 32=[x] < 2*1-\frac 32 = \frac 12$$ so $$[x] =0$$. And $$\{x\} = \frac 34$$ and $$x = \frac 34$$.

Solution 3: $$x = \frac 34$$ and $$[\frac 34] + [ 1\frac 12] + [ 2\frac 14] = 0 + 1 + 2 = 3 = 4*\frac 34$$.

If $$x=I+f$$ where $$0\le f<1$$ and $$I$$ is an integer

If $$3f<1,$$

$$4I+4f=I+2I+3I\implies2I=4f, 0\le I<\dfrac23\implies I=0,f=?$$

If $$1\le3f<2$$ and $$2f<1$$

$$4I+4f=I+2I+3I+1\iff4f=2I+1\implies\dfrac43\le2I+1<2$$ no integer value available for $$I$$

If $$1\le2f$$ and $$1\le3f<2$$

$$4I+4f=6I+2,2f=I+1\implies 1\le I+1<\dfrac43,I=0,2f=1$$

Check if $$2\le3f<3$$