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Stupid number trick III (Posted on 2006-01-11) Difficulty: 3 of 5
Pick any whole number greater than 1.

1. Write down all of its proper divisors (including 1 and the number itself)
2. Add the digits of these divisors.
3. Use this sum to repeat steps 1 and 2 until your number does not change.

Must the process terminate?
At what number(s) can the process terminate at?
What numbers <1000 take the most steps to terminate?

See The Solution Submitted by Jer    
Rating: 4.0000 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Proof? | Comment 2 of 11 |

The number of divisors of n < 2 Sqrt(n);

The number of digits in each divisor <= 1 + Log_10 (n);

The maximum value of each digit is 9.

If f(n) performs one iteration on n, then

f(n) < 9 (1+Log(n) ) 2 Sqrt(n)

For n >= 10000, f(n) < 9 (1+Log(n) ) 2 Sqrt(n) < n; The exact boundary is about 7744

Since the sequence converges to 15 for all 1<n<10000 and is strictly decreasing for all values above 10000, we can conclude that it converges to 15 for all n >1..


  Posted by goFish on 2006-01-11 12:23:52
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