No, 49/5. One dp out (d'oh!). Getting careless in my old age. ('r'-'a'+i)*('u'-'a'+i)*('n'-'a'+i)=187 when i=3 ('t'-'a'+i)*('h'-'a'+i)*('e'-'a'+i)*('s'-'a'+i)*('e'-'a'+i)=5703 when i=3 So if i==3, ('g'-'a'+i)*('o'-'a'+i)*('o'-'a'+i)*('g'-'a'+i)*('l'-'a'+i)*('e'-'a'+i)=2294082 ('y'-'a'+i)*('a'-'a'+i)*('h'-'a'+i)*('o'-'a'+i)*('o'-'a'+i)=234090 2294082/234090=9.8, or 98/10, or 49/5.
@xpi0t0s : Yeah, even I didn't get the logic ... :? The following code uses the "encryption" method you mentioned : Code: #include <stdio.h> int main() { char Decoded[50]; unsigned int Value; int delta; Decoded[0]='O'; while(Decoded[0] != '0') { scanf("%s", Decoded); scanf("%d", &delta); Value = 1; for(int i = 0; Decoded[i] != '\0'; ++i) Value *= (Decoded[i]-'a'+delta); printf("%d\n\n", Value); } return 0; } When I run it ... Code: [B]run 3[/B] [COLOR="Red"]7360[/COLOR] [B]these 3[/B] [COLOR="Red"]226380[/COLOR] [B]google 3[/B] [COLOR="SeaGreen"]2294082[/COLOR] [B]yahoo 3[/B] [COLOR="SeaGreen"]234090[/COLOR] The values in red, don't match with 187 and 5703 respectively. Those is green match perfectly with the values you predict as per your encryption scheme. And, 7360 / 226380 == 368 / 11319 in the simplest form. That might be close to 187/5703 but not equal to it.
yeah I'm way out. Anyway I'm sticking with my guess of 49/50. How about posting your own guess instead of criticising mine.