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Self-numbers and generators (Posted on 2018-05-23) Difficulty: 4 of 5
begin background
For any positive integer n, define d(n) to be n plus the sum of the digits of n.
For example, d(79) = 79 + 7 + 9 = 95.
Take integer n as a starting point, you can construct the infinite increasing sequence of integers n, d(n), d(d(n)), d(d(d(n))), ....etc
For example, if you start with 33, the next number is 33 + 3 + 3 = 39, the next
is 39 + 3 + 9 = 51, the next is 51 + 5 + 1 = 57, and so you generate the sequence : 33, 39, 51, 57, 69, 84, 96, 111, 114, 120, 123, 129, 141, ...
The number n is called a generator of d(n). In the sequence above, 33 is a generator of 39, 39 is a generator of 51, 51 is a generator of 57, and so on. Some numbers have more than one generator: for example, 101 has two generators, 91 and 100.
A number with no generators is a self-number.

There are thirteen self-numbers less than 100: 1, 3, 5, 7, 9, 20, 31, 42, 53, 64, 75, 86, and 97.

end background

Now come my questions:

a.(d2) What is the smallest n making a 10n a self number?

b.(d4) Checking integers with 1 in the beginning, 1 in the end and n-1 zeros between the ones (i.e.10n+1) what value of n creates a number with 3 generators ?

c.(d5 or d4 after a hint) What is the smallest number with 4 generators ?

No Solution Yet Submitted by Ady TZIDON    
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Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): Thoughts on c. (assist req'd) ====>Hint | Comment 7 of 8 |
(In reply to re: Thoughts on c. (assist req'd) ====>Hint by Ady TZIDON)

How is this a hint?  More like a spoiler.  It looks like you've given away the game.

10^24+102 is the answer to part c.

I made the error of presuming the number would be of the form 10^n+1 the way it was in part b.

You should have just pointed this out.  

  Posted by Jer on 2018-05-26 22:31:55
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