Hi Otared and Wolgang,
Thanks for the answers. In fact, being familiar with Python and not at all
with Lua, I didn't have the reflex to think of using a mathematical
function which gives the remainder in the Euclidean division by 10, it's
strange !!
Fabrice

Le jeu. 24 août 2023 à 04:06, Wolfgang Schuster <
wolfgang.schuster.lists@gmail.com> a écrit :

> Otared Kavian schrieb am 24.08.2023 um 01:04:
>
> Hi Fabrice,
>
> As Wolfgang points out, it is indeed possible to fill-in your table with
> Lua: maybe you were wondering how to fill the columns 2, 3 and 4. In this
> case you need to use the Lua function math.mod as in the following, which
> is a completed version of what Wolfgang sent:
>
>
> I guess I completely missed the units digit part :)
>
> %% begin filling-with-lua.tex
> \starttext
>
> \startluacode
> context.startxtable{ align = "middle,lohi", bodyfont = "9pt", framecolor =
> "black" }
>     context.startxrow{ background = "color", backgroundcolor = "lightgray"
> }
>         context.startxcell{ nx = 4 }
>             context("Units digit of")
>         context.stopxcell()
>     context.stopxrow()
>     context.startxrow()
>         context.startxcell{ width = "1cm" }
>             context.im("a")
>         context.stopxcell()
>         context.startxcell{ width = "1cm" }
>             context.im("a^2")
>         context.stopxcell()
>         context.startxcell{ width = "1cm" }
>             context.im("b")
>         context.stopxcell()
>         context.startxcell{ width = "1cm" }
>             context.im("2b^2")
>         context.stopxcell()
>     context.stopxrow()
>     for i = 0,9 do
>         context.startxrow()
>             for j = 1,4 do
>                 context.startxcell()
>                     if j == 1 then
>                         context(i)
>                     elseif j == 2 then
>                     context(math.mod(i*i,10))
>  elseif j == 3 then
>   context(i)
>  else
>   context(math.mod(2*i*i,10))
>                     end
>                 context.stopxcell()
>             end
>         context.stopxrow()
>     end
>
>
> I would drop the column check in this case.
>
>     for i = 0,9 do
>         context.startxrow()
>             context.startxcell()
>                 context(i)
>             context.stopxcell()
>             context.startxcell()
>                 context(math.mod(i*i,10))
>             context.stopxcell()
>             context.startxcell()
>                 context(i)
>             context.stopxcell()
>             context.startxcell()
>                 context(math.mod(2*i*i,10))
>             context.stopxcell()
>         context.stopxrow()
>     end
>
> Wolfgang
>
>
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