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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(3): Going further (520,000,000+) | Comment 12 of 23 |
(In reply to re(2): Going further (520,000,000+) by Justin)

Justin, C++ is OK. 

The Circular Fun problem (and any problem for that matter): You yourself have the power to decide when the solution goes public.  If you feel like now is the right tuime, then log in, go to your problems and then to the problem we're talking about.  Go to the solution and hit the button 'make the solution public' (Or something like that).


  Posted by Hugo on 2006-01-20 12:54:45
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