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I want to solve equations of this type:

$$\lfloor1x\rfloor+\lfloor2x\rfloor+\lfloor3x\rfloor+\lfloor4x\rfloor+\lfloor5x\rfloor=10$$

Vivek Sivaramakrishnan
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    Good for you. To that end, what have you tried? And did you have a question? – amWhy Aug 04 '18 at 14:14
  • I do not know where to start at all. Adding them seems terrible, and the only way I can think is brute-forcing it... – Vivek Sivaramakrishnan Aug 04 '18 at 14:16
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    Hint: $x-1<\lfloor x \rfloor\leq x$. We can estimate $x$. – Jakobian Aug 04 '18 at 14:22
  • If nothing else you can narrow the range of possible solutions, as @Adam suggests. Note that the left hand side is a (weakly) increasing function of $x$. E.g. since setting $x=10$ gives a value that is too large, any solution must be... less than $10$. – hardmath Aug 04 '18 at 14:32
  • @hardmath $x = 1$ is already too large. – Theo Bendit Aug 04 '18 at 14:45
  • Does anyone know of any techniques besides brute force computation? I can tell you it's probably more than 0.5, but less than 1. – CaptainAmerica16 Aug 04 '18 at 14:47
  • @CaptainAmerica16 now go one step ahead just check it for 3/4 which gives 9 thus the solution is between 0.75 and 1 – Deepesh Meena Aug 04 '18 at 14:58
  • @CaptainAmerica16 I followed your approach using the gradient descent techinque and get 0.8, sometimes 0.9 by assuming the cost as the output of the function and gradient as 10 minus the cost. – Vivek Sivaramakrishnan Aug 04 '18 at 15:03

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Think of it like there are five rotating spinners, where the $i^{th}$ spinner makes $i$ revolutions per minute. All spinners start at the same position at $x=0$ minutes. The first spinner completes a revolution every $60$ seconds, the second every $30$ seconds, etc. The number of revolutions completed at time $x$ by the $i^{th}$ spinner is $\lfloor ix\rfloor$, so your problem is to find all of the times where the combined total of revolutions is $10.$

Make a schedule of all the times a revolution is completed (a "tick"):

Spinner 1 ...........................................................x
Spinner 2 .............................x.............................x
Spinner 3 ...................x...................x...................x
Spinner 4 ..............x..............x..............x..............x
Spinner 5 ...........x...........x...........x...........x...........x

Each . is one second, and each x is a tick. We see that the $10^{th}$ tick occurs when Spinner 5 makes its penultimate tick at $x=0.8$ minutes $=48$ seconds. Therefore, the set of times for which there are $10$ ticks total is $x\in [0.8,1)$.

Mike Earnest
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You can start from the following.

$x=[x]+\{x\}$ and consider number of cases:

  1. $0\leq\{x\}<\frac{1}{5};$

  2. $\frac{1}{5}\leq\{x\}<\frac{2}{5};$

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$.$

$.$

Good luck!

Also, the left side increases!

Michael Rozenberg
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  • Wouldn't you want a finer partition than that? The parts in this partition will make $\lfloor 4x \rfloor$ annoying to keep track of, for example. – Theo Bendit Aug 04 '18 at 14:38
  • @Theo Bendit It's the standard notation. $[x]$ and $\lfloor {x}\rfloor$ they are the same. There is also $\lceil x\rceil$ of course, but it's another story. If you say about the essence of my hint then it's the hint only. I did not write all cases. – Michael Rozenberg Aug 04 '18 at 14:44
  • @MichaelRozenberg $ \lfloor x \rfloor $ to me means floor function where I have seen $ [x] $ as just being used as brackets, equivalent to $ (x) $. – Warren Hill Aug 04 '18 at 14:50
  • @Warren Hill We always know from the context about which thing we say. – Michael Rozenberg Aug 04 '18 at 14:52