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 Find From Floor (Posted on 2015-01-30)
Find all possible values of a real number x that satisfy this equation:
```9x       floor(x)
---- = --------------
10      x – floor(x)```

 No Solution Yet Submitted by K Sengupta No Rating

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 solution | Comment 1 of 4
From the graph (click to see), one looks for solutions between 1 and 2, between 2 and 3, etc., probably ending around 10 or 11.

First one:

9(1+f) / 10 = 1/f

where f is the fractional part.

9 + 9f = 10/f
9f + 9f^2 -10 = 0

f = (-9+sqrt(81+360))/18
= 2/3

first solution: 5/3

Then

9(2+f) / 10 = 2/f
18 + 9f = 20/f
18f + 9f^2 -20 = 0
9f^2 + 18f - 20 = 0

f = (-18 + sqrt(324+720))/18
~= .7950549357115

second solution: approx. 2.7950549357115

In general

k + (sqrt(81k^2 + 360k) - 9k)/18

The valid values of k are 1 through 8, as the table shows:

1             1.666666666666667
2             2.795054935711502
3             3.862907813126304
4             4.905932629027116
5             5.935921354681383
6             6.95811402901264
7             7.975240527365851
8             8.988876515698589
9             10
10            11.00925212577332
11            12.01707159867238
12            13.02376916856849
13            14.02957133205101
14            15.03464719546263
15            16.03912563829967
16            17.04310664416703

as 10 is not valid as 9.99999... in this context, i.e., floor(10) <> 9; and, the more so, floor(11+) <> 10, etc.

table from:

DEFDBL A-Z
FOR k = 1 TO 16
PRINT k,
f = (SQR(81 * k * k + 360 * k) - 9 * k) / 18
PRINT k + f
NEXT

 Posted by Charlie on 2015-01-30 14:36:09

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