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CESS.tk wrote:
Here's the solution to (B):[pre] 1 = one 1 = 11
11 = two 1's = 21
21 = one 2, one 1 = 1211
1211 = one 1, one 2, two 1's = 111221
111221 = three 1's, two 2's, one 1 = 312211
312211 = one 3, one 1, two 2's, 2 one's = 13112221
13112221 = one 1, one 3, two 1's, three 2's, one 1[/pre]So the correct answer is 1113213211
With this information, riddle (A) shouldn't be too hard anymore.
1311221113...
but I wish I had made it before you told the above...
I'm afraid your answer is incorrect. The technique used in riddle (A) is similar to riddle (B), but not the same. Try to find out what's happening in this series:
1
11
21
1112
3112
211213
312213
Notice the first three numbers are the same as in riddle (B), but the fourth is different.