Var
  i,k,t: longint;

Begin
    readln (input,k);
    t:= round(sqrt(k)) +1;
    for i:= 3 to t do
        if (k mod i) = 0 then break;
    If i<>t then writeln (output, i-1)
            else begin
                 if ((k mod 2) = 0) then k:= k shr 1;
                 writeln (output, k-1);
                 end;
End.

> Var
>   i,k,t: longint;
>
> Begin
>     readln (input,k);
>     t:= round(sqrt(k)) +1;
>     for i:= 3 to t do
>         if (k mod i) = 0 then break;
>     If i<>t then writeln (output, i-1)
>             else begin
>                  if ((k mod 2) = 0) then k:= k shr 1;
>                  writeln (output, k-1);
>                  end;
> End.
>

(I had already checked that case but then I change a little the code
and forgot about it.
 I solve that problem by changing
t:= round(sqrt(k)) +1; to
t:= round(sqrt(k)) +3;
 bt I still get "wrong answer". Any other ideas?
  By the way, &#1111;from which state are you?)


 Ya hab&#1085;a checado ese caso pero lo cambie un poco el c&#1091;digo y no lo
volv&#1085; a revisar.
 Arregle ese problema cambiando
t:= round(sqrt(k)) +1; por
t:= round(sqrt(k)) +3;
 pero a&#1098;n asi no me lo acepta. &#1111;Tienes otra idea?
  A prop&#1091;sito, &#1111;de que &#1081;stado eres?

I think your solution is OK, but you need to add the special case of
k = 4, (with an if). And i'm from Mty, i'm studying at ITESM. Where
are you?. Good Luck.
mail: miguelangelhdz@hotmail.com