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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.)
Some Thoughts part a. with positive n | Comment 2 of 8 |
10=5+5
100=86+8+6
1000=977+9+7+7
10000=9975+9+9+7+5
100000=99959+9+9+9+5+9
The answer is n=6
Proof that 10^6=1000000 is a self-number

Assume 1000000 has a generator
the generator must be below 999990, since 9*5>10
the generator must be above 999899, since 9*5+5<101
so the generator is of the form 9999ab for digits a, b
d(9999ab)=999900 + 10a + b + 36 + a + b
so 
11a+2b=64
there are no digits from 0 to 9 that solve this equation, therefore 10^6 has no generator.

  Posted by Jer on 2018-05-23 14:25:31
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