Two players (say A and B) take turns, starting as follows"
ABBABAABBAABABBA....
(they alternate for the first two. Then they alternate, but with B first. For the next four, the do the first four, but reversing A and B. For the next eight, they do the first eight, but reversing A and B. Etc.)
Without writing out the sequence, can you determine who takes the 99th and 999th turn?
(HINT: IT HAS SOMETHING TO DO W/BINARY NUMBERS)
2007-02-22 08:36:26 · 3 answers · asked by ClooneyIsAGenius 2 in Computers & Internet ➔ Programming & Design
Hmmm....very good question, Im not sure at the moment. (It doesnt repeat every 12 though).
Checked back to see if there was an answer - How does this deal with fibonacci numbers? They go
1,1,2,3,5,8,13....
I dont see that pattern anywhere. You arent adding up the 2 previous terms, your adding up all the previous terms each iteration.
2007-02-22 09:08:50 · answer #1 · answered by rsmith985 3 · 1⤊ 0⤋
Hehe, there is a pattern, quite a famous one too. Count the number of terms for each iteration in the sequence. The values for each iteration are the Fibbonnacci numbers. Knowing that, you only need to calcuate F(99) and F(999) to get your answers.
2007-02-23 10:03:50 · answer #2 · answered by Pfo 7 · 0⤊ 0⤋
Not sure how this has to do with binary numbers but
the pattern is ABBABAABBAAB --repeat which is 12 do do an integer division...
999/12 = 83 R 3.... so the answer is B
I don't see how this is related to binary numbers... ... I would be interested in finding out...
2007-02-22 16:51:09 · answer #3 · answered by AvidBeerDrinker 3 · 0⤊ 1⤋