I made a C program to solve the problem. The answer is

25 18 25 24 23 24 26 25 26

Here's the program:

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#include <stdio.h>

#define DEBUG 0

// Return number of digits other than digit b in number a
int checkDigits(int a,int b)
{
 int ans = 0;

 do {
  if (a % 10 != b)
   ans++;
 } while ( (a = a/10) != 0);

 return ans;
}

int main()
{
 int i, j, k, n, ok;
 int a[9]; // Sequence from 1 to 9
 int b[9]; // Answer
 int c[9]; // Backup of b
 int d[9]; // Buffer

 // Initialize vectors 
 for (i=1;i<=9;i++)
 {
  a[i-1] = i;
  b[i-1] = 0;
  c[i-1] = 0;
  d[i-1] = 0;
 }

 while (1)
 {
  // Loop through all 9 digits
  for (j=1;j<=9;j++)
  {
   // Count instances of digits other than j 
   k = 0; 
   for (i=1;i<=9;i++)
   {
    if (a[i-1] != j) k++; 
    if (b[i-1] != 0 && (n = checkDigits(b[i-1],j)) )
     k += n;
   }
   // Save result in buffer
   d[j-1] = k;
  }

  // Transfer buffer to b for next iteration
  for (i=0;i<9;i++)
   b[i] = d[i];

  ok = 0; // Set to 1 if answer is the same as after last iteration
  for (i=0;i<9;i++)
  {
   if (b[i] != c[i])
    break;
   if (i==8) ok = 1;
  }

  #if DEBUG == 1
  for (i=0;i<9;i++)
   printf("%i %i %i\n",a[i],b[i],c[i]);
  printf("\n");
  #endif

  if (ok) // Break if answer hasn't changed 
   break;
  else // Save answer if it has changed 
   for (i=0;i<9;i++)
    c[i] = b[i];
 }

 // Print results
 for (i=0;i<9;i++)
  printf("%i %i\n",a[i],b[i]);

 return 0;
} 

For starters, let's first recognize that the number of digits is going to be at least 18, so if each of these answers is going to be double digits, the total number of digits will come out to be 9*3 = 27.

Next, we find that excluding 2, almost every digit is not going to appear very often, so all of the answers with the exception of 2 are going to be close to and at most 26. This means that the number of digits that are not 2 is going to be close to and at most 18.

This next bit helps optimize the counting a little bit. Since the total number of digits is 27, rather than counting up the number of digits that are not a particular one, we count up the number of times a digit appears and subtract that from 27. So if the digit 8 appears twice, then the count of non-8 digits is 27-2 = 25.

The digits 7 and 9 are only going to ever appear once, making the count 26. The digit 8 will probably only appear twice because of the 18 for answer 2. Similarly, the digit 1 will probably only appear twice. Both of these counts is 25.

This leaves us with the following likely scenario:

1: 25

2: 18

3: 2-

4: 2-

5: 2-

6: 2-

7: 26

8: 25

9: 26

The last 4 answers are very erratic, as changing one of them usually changes many of the others. However, we can now prove that the numbers we have fixed will not be changed by just playing with the missing digits.

At most, digits 3 and 4 can have 5 occurrences, but if one of them did have 5, the count comes to 27-5 = 22, which does not contribute the digit needed. Therefore, they can only have 4 occurrences maximum, which brings the count of a digit to 23 at the lowest, which does not change any of the other quantities we have. The only digits that can extend past 4 occurrences is 5 or 6, since they already appear three times each, meaning they start at 24, so it is not possible for them to appear more than 4 times either. (There are one or two ambiguous cases, but working through them manually shows they don't work.)

However, just proving that these missing digits will not affect our fixed digits does not mean the answer exists. Between the possibilities of 3, 4, 5, and 6, there are 4^4 = 256 possible cases to check, in which the large majority of them can be immediately or quickly discarded. This now becomes possible to guess and check by hand, which allows us to find our final answer.

I thought this would be much harder than the easier counterpart, but to my surprise only 11 trials is required to solve this puzzle. Nevertheless, I've written a program to solve this problem: link .

For those who can't read code or can't be bothered to click the link, here are some explanations. Some answers have already proved that all the answers must be 2 digits long and there are maximum 27 digits in the box, so my program simply starts by placing 27 on all the underlines and start counting the number of digits in the box, then simply place the new answers as 27 minus the number of digits and repeat the process.

Here are the all the trials needed (so doing it by hand is probably not that hard either):

• 261726262626172626
• 241926262619242626
• 241926242621262624
• 241826232622262625
• 251725252522262526
• 251726262124252626
• 241826252422252626
• 251726242423262526
• 251825242423252626
• 251825242324262526
• 251825242324262526

Alternative solution: eighteighteighteighteighteighteighteighteight.

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"Complementary Counting"
def solve():
    ll = [8] *9
    ss = "".join(map(str, ll))
    tt = ""
    while tt != ss:
        ll = [8 + len(ss.replace(str(k + 1),'')) for k in range(9)]
        tt, ss = ss, "".join(map(str, ll))
    return ss

print(solve())

Output :

1

'251825242324262526'

Program

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prev = None
step = []
while prev != step:
    print(step) # shows the different steps
    prev = step
    step = [8] * 9
    for cnt in prev:
        for d in str(cnt):
            step = [i + 1 for i in step]
            step[int(d) - 1] -= 1
print(''.join(map(str, step)))

Output

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[]
[8, 8, 8, 8, 8, 8, 8, 8, 8]
[17, 17, 17, 17, 17, 17, 17, 8, 17]
[17, 25, 25, 25, 25, 25, 17, 24, 25]
[24, 19, 26, 25, 20, 26, 24, 26, 26]
[25, 18, 26, 24, 25, 22, 26, 26, 24]
[25, 17, 26, 24, 24, 23, 26, 25, 26]
[25, 18, 25, 24, 24, 23, 25, 26, 26]
[25, 18, 25, 24, 23, 24, 26, 25, 26]
251825242324262526

As we're submitting programmable solutions, I may as well add an alternative which I achieved in LibreOffice Calc, with Iterations enabled in Tools - Options - LibreOffice Calc - Calculate.

• Create 1-9 in cells A1:A9.
• Copy the first formula from the table below.
• Paste into B1.
• Copy B1.
• Paste into B2:B9.
• Keep pasting back into B1 until the numbers stop iterating.

You have to include all digits in the box in your counts. This includes the digits in your answer. This means the answers will all be two digits. 25 18 25 24 23 24 26 25 26

251825242324262526

You have to include all digits in the box in your counts. This includes the digits in your answer. This means the answers will all be two digits.

1. 25
2. 18
3. 25
4. 24
5. 23
6. 24
7. 26
8. 25
9. 26