# Equation involving floor function without real solutions

Prove that equation $$\lfloor x\rfloor\;+\;\lfloor 2x\rfloor\;+\;\lfloor 4x\rfloor\;+\;\lfloor16x\rfloor\;+\;\lfloor32x\rfloor\;=12345$$ has no real solutions.

Edit:

My thoughts from $$5$$ months ago (from the comment) that should've been included in the post:

I first checked whether there are integers, then I tried to see how the decimal part changes when multiplying, because the change might be more than multiple of the previous term... and I also included the coefficients.

• What have you tried? – Viktor Glombik Nov 11 '19 at 7:39
• @ViktorGlombik I first checked whether there are integers, then I tried to see how the decimal part changes when multiplying, because the change might be more than multiple of the previous term... and I also added the coefficients. – Fractal Nov 11 '19 at 8:01
• Hint: $12345 = 224 \times 55 + 25$ – Gribouillis Nov 11 '19 at 8:05
• Thanks! I'll experiment. – Fractal Nov 11 '19 at 8:06

Let $$x=n+r$$ with $$n$$ an integer and $$0\leq r <1$$. Then your equation reduces to

$$55n + \lfloor r \rfloor + \lfloor 2r \rfloor + \lfloor 4r \rfloor + \lfloor 16r \rfloor +\lfloor 32r \rfloor = 12345.$$

We have $$\lfloor r \rfloor + \lfloor 2r \rfloor + \lfloor 4r \rfloor + \lfloor 16r \rfloor +\lfloor 32r \rfloor \leq 0 + 1 + 3 + 15 + 31 = 50.$$ The only multiple of $$55$$ within $$55$$ of $$12345$$ is $$12320$$, so we must have

$$\lfloor r \rfloor + \lfloor 2r \rfloor + \lfloor 4r \rfloor + \lfloor 16r \rfloor +\lfloor 32r \rfloor = 25.$$

But if $$r\geq 1/2$$, the left side is at least $$27$$. And if $$r<1/2$$ the left side is less than $$23.$$

• How would you know you need to test for $r = \frac{1}{2}$? – Toby Mak Nov 12 '19 at 0:12
• @TobyMak First, I see all the powers of two. The first term is zero, so I look at the second term. It's zero for $r<1/2$ and $1$ for $r>1/2$. So I split into two cases. My plan was to continue splitting, but it turned out that the first two cases covered it. – B. Goddard Nov 12 '19 at 2:58

Too long for a comment.

I assume you are not allowed access to a calculator. From Griboullis's hint, you know that if $$x$$ exists, $$224 ≤ x ≤ 225$$, so the only thing to do is to find the fractional part, which is given by $$\lfloor x\rfloor\;+\;\lfloor 2x\rfloor\;+\;\lfloor 4x\rfloor\;+\;\lfloor16x\rfloor\;+\;\lfloor32x\rfloor\;=25$$.

Then prove that the function jumps by one unit whenever $$x$$ is only a multiple of $$\frac{1}{32}$$, by two units whenever $$x$$ is only a multiple of $$\frac{1}{16}$$, and you can work out for the rest up to a jump of four units whenever $$x$$ is a multiple of $$\frac{1}{2}$$. Adding up all these jumps methodically should lead you to the point $$(0.499, 23)$$ (in the simplified question), at which point the next jump gives $$(0.5, 27)$$. Transforming the question back to the original question will complete the proof

$$y-1<[y]\le y$$ for all real $$y$$, so $$55x-5<[x]+[2x]+[4x]+[16x]+[32x]\le55x$$ so $$55x-5<12345\le55x$$, so $$12345/55\le x<12350/55$$, that is, $$224.454545\dots\le x<224.545454\dots$$; better, $$224{5\over11}\le x<224{6\over11}$$. When $$x=224{5\over11}$$, we get $$[x]+[2x]+[4x]+[16x]+[32x]=55\times224+[5/11]+[10/11]+[20/11]+[80/11]+[160/11]=12320+0+0+1+7+14=12342$$ When $$x=224{6\over11}$$, we get $$[x]+[2x]+[4x]+[16x]+[32x]=55\times224+[6/11]+[12/11]+[24/11]+[96/11]+[192/11]=12320+0+1+2+8+17=12348$$ Now $$5/11=(14.5\dots)/32$$, and $$6/11=(17.4\dots)/32$$, so we only have to look at what happens when we move past $$15/32$$, $$16/32$$, and $$17/32$$. Moving past $$15/32$$, the sum goes up by one, to 12343. Moving past $$16/32=1/2$$, the sum goes up by four, to $$12347$$, and we've gone past $$12345$$.