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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!

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    $\begingroup$ Drawing the graph might give you some ideas of how to attack the problem: wolframalpha.com/input/… $\endgroup$ – Hans Lundmark Jul 21 '19 at 18:12
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    $\begingroup$ 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 $\endgroup$ – Jakobian Jul 21 '19 at 18:14
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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 $$\overline{x}=\frac{0+\frac12+\frac34}3=\frac5{12}$$

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  • $\begingroup$ 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! $\endgroup$ – Falcon Rhymes Jul 21 '19 at 18:36
  • $\begingroup$ Done. Thanks again! $\endgroup$ – Falcon Rhymes Jul 21 '19 at 19:10
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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 $$3=2k+1+2 \Rightarrow x=\frac{3}{4} $$
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Easy answer:

$[x] + [2x] + [3x] = 4x$ 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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Hard answer:

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

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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 $2\le3f<3$

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