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1
11
21
1211
111221
312211
13112221
1112213211
312211131221
1311223113112211
11132122132113212221
3113121122111312211312113211

2007-06-07 10:02:26 · 6 answers · asked by shadysbr0ken 1 in Science & Mathematics Mathematics

6 answers

13211311122122311311222113111221131221.




Look at the top of the list [1]. What is the number?
There is one (1) 'one' (1)
So the next number of the series is 11.

Look at the next number [11].
There are two (2) 'one's (1)
So the next number of the series is 21

Look at the next number [21].
There is one (1) 'two' (2) and one (1) 'one' (1)
So the next number of the series is 1211

Look at the next number [1211].
There is one (1) 'one' (1), one (1) 'two' (2) and two (2) 'one's (1)
So the next number of the series is 111221

Look at the next number [111221].
There are three (3) 'one's (1), two (2) 'two' (2) and one (1) 'one' (1)
So the next number of the series is 312211



Following this pattern, you will get your answer.


Interestingly, the numbers will never have a '4' in them (max is 3).

2007-06-09 03:42:32 · answer #1 · answered by to0pid 2 · 0 0

1 3 2 1 1 3 1 1 1 2 2 1 2 2 3 1 1 3 1 1 2 2 2 1 1 3 1 1 1 2 2 1 1 3 1 2 2 1

Each line describes the one before it.
eg in the 2nd line, there are 2 1's, so the third line is 2 1 etc.

2007-06-08 23:32:03 · answer #2 · answered by Dr D 7 · 0 0

Each new number "describes" the preceding number,
so that for example 312211 consists of
one 3, one 1, two 2s, two 1s,
so its successor is 13112221.

If so, there is an error in the list: the eighth term should be
1113213211.

But the rule continues after that, so the "next" number is
13211311122122311311222113111221131221.

2007-06-07 17:26:32 · answer #3 · answered by David Y 5 · 0 0

132113111221223113112221
13111221131221

(one number continued on the second line)

2007-06-07 17:26:05 · answer #4 · answered by Anonymous · 0 0

im stumped, is this a legit question?

2007-06-07 17:10:21 · answer #5 · answered by Steve C 2 · 0 0

31131211221113122113121132113

2007-06-07 17:28:28 · answer #6 · answered by Anonymous · 0 0

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