[Jprogramming] Math Problem

[Skip Cave]

I made a mistake in the equation in my first post..

The three terms that are multiplied are
1. x
2. floor of x  = (<.x)
3. fractional part of x = (x - <.x) This is what I got wrong in my first
post.

I can get close by manual trial & error:

*x=.64.962573478*

Floor of x = <.x = 64

Fractional part of x = (x -<.x ) = 0.962573478

* x: x * (<.x) * (x - <.x)*

*213746821r53410*


Close, but no cigar.


* 4002 = x * (<.x) * (x - <.x)*

*0*

A closer look - as a decimal fraction:

* x * (<.x) * (x-<.x)*

*4002.0000187231198652*


Yep. Not close enough.


How to design an iterative solution? There should be multiple solutions
with (<.x) = 63, 64, 65. 66 ... with the fraction

*(x-<.x) *getting smaller & smaller.

Skip Cave
Cave Consulting LLC


On Mon, Oct 18, 2021 at 6:25 PM Skip Cave <skip at caveconsulting.com> wrote:

> How to solve this problem?
>
> 4002x = n * (<.n) * (>.n)
>
>
> What is n, where n is a rational fraction greater than 1, and the answer
> is a rational fraction? There are likely many answers, so find some answers
> near 64. The result in J should be a 1:
>
> 4002x = n * (<.n) * (>.n)
> 1
>
>
> Skip
>
> Skip Cave
> Cave Consulting LLC
>