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Re:Riddles
Posted: Mon Dec 30, 2002 11:07 pm
by SGreenslade
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...
Re:Riddles
Posted: Tue Dec 31, 2002 2:59 am
by CESS.tk
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.
Re:Riddles
Posted: Tue Dec 31, 2002 6:00 am
by David
212223
114213
31121314
41122314
31221324
21322314
21322314
and so it ends...
Re:Riddles
Posted: Tue Dec 31, 2002 6:30 am
by CESS.tk
David wrote:
212223
114213
31121314
41122314
31221324
21322314
21322314
and so it ends...