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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.)
interesting property | Comment 2 of 23 |

The process basically goes from odd number to odd number.  If we break the odd numbers up into two groups:

1+4n and 3+4n where n=0,1,2,....

then every other odd number will result in a smaller odd value

(3(1+4n)+1)/2^m = (4+12n)/2^m = 1+3n (m=2)

and 1+3n < 1+4n for all n

Now for the other half of the odd numbers

(3(3+4n)+1)/2^m = (10+12n)/2^m = 5+6n (m=1)

For all n, 5+6n is odd and the step ends and is > 3+4n

So alternating odd numbers either diverge from or converge toward 1, the smallest odd number.  For the process not to converge, it would have to either hit mostly divergent values or find a cycle.


  Posted by Bob Smith on 2006-01-13 15:36:31
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