i can't find my bug.
give me test#48, thx.

    var
     x, n : integer;
     t : real;

begin
  readLn(n);
  if n mod 10 = 7 then t := (ln(3)+(n - 2) * ln(2))/ln(10)
  else t := (ln(3)+(n-1) * ln(2))/ln(10);
  x := TRUNC(t);
  t := t - x;
  t := exp(t*ln(10));
  if t < 1.5 then dec(x);
  writeLn(x);
end.

Edited by author 05.03.2008 22:39

Could you explain your program ?
How you get this formulas ?

Thank you in advance !

if n is very big number then a[n] will be 2/3;
lim(a[n]) = 2/3 when n->MAXnumber;
you must continued fo'rmula, then you will have a[n] := (...)/2^(n-1);
you must write this with 10^x (x-will be answer);
like this
10^x = 2^(n-1);
x =lg(2^(n-1))  =>  x = ln(2^(n-1))/ln(10);

Hm, n ≤ 100000, but I've always thought that Integer in Free Pascal ≤ 32767...
Anyway, I recommend you to switch Integer to Longint and Real to Extended.

it doesn't work...

t := (ln(3)+(n-1) * ln(2))/ln(10) ...
<- its really so

if n mod 10 = 7 then t := (ln(3)+(n - 2) * ln(2))/ln(10) ...
<- but i cant understand this rule

if t < 1.5 then dec(x) ...
<- is only?

you realy must compare:

 +2/3/(2^n)  and  1/3/(10^t)  - when odd(n)

 -2/3/(2^n)  and  -2/3/(10^t) - when even(n)

why its so? because now you must know will the remainder (+/-) 2/3/(2^n)  corrupt the (t-1)-th digit of your diabolic number

Edited by author 11.03.2008 22:18

thanks a lot! with your help, now i've AC