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Stupid number trick IV (Posted on 2006-01-13) Difficulty: 3 of 5
Take any whole number greater than one.

1. If the number is odd, multiply it by three and add one.
2. If it is even divide it by two.
2a. If the result is still even, continue to divide by two until the result is odd.
3. Continue steps 1 and 2 until you get the same number twice.

[For example starting with 9 -> 28 -> 14 -> 7 which is considered one iteration. The next iteration brings this to 11.]

What number(s) does this process terminate at?
What starting value less than 200 takes the most iterations to terminate?

See The Solution Submitted by Jer    
Rating: 4.0000 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: Collatz Conjecture | Comment 7 of 23 |
(In reply to Collatz Conjecture by The riddler)

"The process terminates at 4->2->1->4..." ??

If you check the definition of an iteration you will see that all the terms in the process (other than the first) must be an odd number.  So all the sequences (we conjecture) end ...1, 1.


  Posted by goFish on 2006-01-14 02:20:05
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