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back to boardThis 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 |
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