This is my programm;
Compil. PASCAL 7.0

[code deleted]

Edited by moderator 13.02.2007 20:47

Try next tests by my Ac prog It may be ny prog also wrong
if Ok but WA5 I send additional

4 !!!!
4
7 !!
105
12 !!!!!!!!
48
19 !!!!!
9576
24 !!!!!!!!!!
1344

Why

12 !!!!!!!! = 48

k:=8;
   12*(12-8)*(12 mod 8)=12*4*4
12 mod 8=4;

19 !!!!! =9576

k:=5;
19*(19-5)*(19-10)*(19-15)*(19 mod 5)=19*14*9*4*4=38304
19 mod 5=4;

12 !!!!!!!!
k=8
12 mod 8=4
12*(12-8)*(12-2*8)...(12 mod 8)=12*4=48
(12 mod 8) is last number from sequence n(n-k)....(n mod k)
but not additional multiplyer as you think
PS! I think that expierenced members of the syte should
send your training test much bore broader than now
for reason that begginers trying too many time on adjusting
their mind to problems understanding

what does ... mean then?  how can it not be a multiplyer?  also (12-2*8) is negative?  I assumed the sequence stopped before 12-ik became negative.

Can someone explain this problem to me in a human language?  Please?

12 !!!!!!!! = 12*12mod8 (not 12*(12-8)*12mod8 'cause 12-8 = 12mod8)

Edited by author 15.12.2007 23:48

Why n = 12 if n <= 10?