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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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Some Thoughts Thoughts on c. (assist req'd) | Comment 5 of 8 |
I'm not quite stuck, but it's really difficult to continue.  So I'm sharing my progress to see what others can do.

Note:  Two or more letters or numbers with no symbol between them stands for concatenation so 9BA means 900+10B+A.  If I want multiplication I'll use the symbol *.

He's a way of looking at part b.
10^n+1 is clearly generated by 10^n since the sod=1
(I will henceforth ignore this and look for two or three additional generators.)

Other generators must start with a bunch of 9's
for example for n=6 d(999955)=1000001 

If we seek to make just the last two digits non-9 we won't be able to find 10^n-1 with two generators.
Suppose for n=7 we seek digits B, A such that 
d(99999BA)=10000001
99999BA+9+9+9+9+9+B+A=10000001
11B+2A+9*5=10000001-9999900=101
11B+2A=101-45=56
This linear combination of B and A with both valid digits will never have two solutions (check for yourself) (and sometimes has none)  The only solution, in this case, is B=4, A=6.

Once we get to the point where a 3rd digit can be non-9 we can get an extra solution.  This happens to be at n=13
d((ten9's)CBA)=10^13+1
gives 
101C+11B+A+90=1001
101C+11B+A=911
here the second solution comes because there are two possible values of C (and each gives a single valid solution)
C=9 makes 11B+2A=2 so B=0, A=1.
C=8 makes 11B+2A=103 so B=9, A=2.
so the final three digits can be 901 or 892  

These triple generators occur rarely and only because 11B+A can slightly exceed 100.  11*9+9=117 is the maximum.  They cannot give any of {1,3,5,7,9,20,31,42,53,64,75,86,97,108,110,112,113,116}

So far so good.  It seems that going to a fourth non-9 digit D should do the trick.  No so!

The greatest of the 81 trios 101C+11B+A that has two sums is 925 (CBA is 908 or 899.)  This is smaller than the minimum contribution of the digit D which is 1001.  

(Incidentally, the smallest number that invokes D is n=112 
d(109*9DCBA)10^113+1 has a single last digit solution 8992.  We are already past a googol.)  

Ok, all is not lost.  Let's use digit E.   The contribution of 10001E+1001D together can have a narrower gap.  (E=9,D=0)-(E=8,D=9)=992.  Better but not small enough.
F then?  Nope.  983.
Going down by 9s.  Ok.  Keep adding digits.
MLKJIHGFED can have a gap as small as 920, so 925 should be possible for CBA.

This is where I stand.

M=9, L=K=J=I=H=G=F=E=D=0 and
M=8, L=K=J=I=H=G=F=E=D=9 both yield 920.  
There actually aren't two CBA solutions for 920 just (CBA=910)

So the number of digits is a number that itself is probably around 13 digits long, they are mostly 9's except the last 13.  It is hard to continue because I'm using a calculator with a 10 digit display.

I'm close though.  Hopefully, this makes enough sense that someone else can follow it through.  At least I've pinned down the generators at less than a googolplex, right?

Edited on May 25, 2018, 10:13 pm
  Posted by Jer on 2018-05-25 22:03:03

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