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

Canada,1981


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.

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    $\begingroup$ What have you tried? $\endgroup$ – Viktor Glombik Nov 11 '19 at 7:39
  • $\begingroup$ @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. $\endgroup$ – Fractal Nov 11 '19 at 8:01
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    $\begingroup$ Hint: $12345 = 224 \times 55 + 25$ $\endgroup$ – Gribouillis Nov 11 '19 at 8:05
  • $\begingroup$ Thanks! I'll experiment. $\endgroup$ – Fractal Nov 11 '19 at 8:06
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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 $23.$

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  • $\begingroup$ How would you know you need to test for $r = \frac{1}{2}$? $\endgroup$ – Toby Mak Nov 12 '19 at 0:12
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    $\begingroup$ @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. $\endgroup$ – B. Goddard Nov 12 '19 at 2:58
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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

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$$y-1<[y]\le y$$ 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 $12345$.

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