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Addition sequence (Posted on 2007-01-09) Difficulty: 2 of 5
1, 1, 2, 11, 12, 13, 112, 113, 114, 1104, 2004, 2012, 1031, 123, ...

Name the next few numbers in the sequence. What is the ultimate fate of this sequence?

See The Solution Submitted by Tristan    
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Solution Puzzle Solution Comment 5 of 5 |
(In reply to Answer by K Sengupta)

Let T(n) = nth term of the sequence.

Then,
Units digit of T(n) = # 1's in T(n-1) and T(n-2) taken together.

Tens digit of T(n) = # 2's in T(n-1) and T(n-2) taken together.

Hundred digit of T(n) = # 3's in T(n-1) and T(n-2) taken together, and so on.

Accordingly, we must have:

T(13) = 1031 (given)
T(14) = 123 (given), so that:
T(15) = 213
T(16) = 222
T(17) = 141

T(18) = 1032
T(19) = 1113
T(20) = 214
T(21) = 1114
T(22) = 2014
T(23) = 2014
T(24) = 2022
T(25) = 1041
T(26) = 1031
T(27) = 1032
T(28) = 213
T(29) = 222
T(30) = 141

-----------
----------, and so on.

Consequently, we can now assert that:

(i) The respective missing 15th term, 16th term and the 17th term are 213, 222 and 141.

(ii) From the 18th term onwards, the given sequence will loop indefinitely.


  Posted by K Sengupta on 2008-12-09 06:32:21
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