aime26-01

Proposed solution

To solve the problem of finding the number of positive integer palindromes written in base 10 with no zero digits, and whose digits add up to 13, we start by understanding the structure of a palindrome. A palindrome reads the same forwards and backwards. For a number to be a palindrome, it must have the form \( \overline{a_1a_2 \ldots a_k a_k \ldots a_2a_1} \) where \( a_i \) are the digits.

Given that the digits must sum to 13 and there are no zero digits, we need to consider the possible lengths of the palindrome and the constraints on the digits.

### Step-by-Step Solution:

1. **Single-digit Palindromes**: 
   - These are not possible since a single digit cannot sum to 13.

2. **Two-digit Palindromes**: 
   - These are not possible since the sum of two identical digits cannot be 13.

3. **Three-digit Palindromes**: 
   - These have the form \( \overline{aba} \). The sum of the digits is \( a + b + a = 2a + b = 13 \).
   - Solving for \( b \), we get \( b = 13 - 2a \). Since \( a \) and \( b \) are digits (1 through 9), we need \( 13 - 2a \) to be a digit (1 through 9).
   - This gives us \( 1 \leq 13 - 2a \leq 9 \), which simplifies to \( 2 \leq 2a \leq 12 \) or \( 1 \leq a \leq 6 \).
   - Possible values for \( a \) are 1, 2, 3, 4, 5, and 6. For each \( a \), \( b = 13 - 2a \) gives us:
     - \( a = 1 \rightarrow b = 11 \) (not valid since \( b \) must be a digit)
     - \( a = 2 \rightarrow b = 9 \) (valid)
     - \( a = 3 \rightarrow b = 7 \) (valid)
     - \( a = 4 \rightarrow b = 5 \) (valid)
     - \( a = 5 \rightarrow b = 3 \) (valid)
     - \( a = 6 \rightarrow b = 1 \) (valid)
   - Valid three-digit palindromes: 297, 585, 753, 831, 951.

4. **Four-digit Palindromes**: 
   - These have the form \( \overline{abba} \). The sum of the digits is \( 2a + 2b = 13 \).
   - Solving for \( b \), we get \( b = \frac{13 - 2a}{2} \). Both \( a \) and \( b \) must be digits (1 through 9).
   - This gives us \( 13 - 2a \) must be even and between 2 and 18.
   - Possible values for \( a \) are 1, 2, 3, 4, 5, and 6. For each \( a \), \( b = \frac{13 - 2a}{2} \) gives us:
     - \( a = 1 \rightarrow b = 5.5 \) (not valid)
     - \( a = 2 \rightarrow b = 5 \) (valid)
     - \( a = 3 \rightarrow b = 3.5 \) (not valid)
     - \( a = 4 \rightarrow b = 3 \) (valid)
     - \( a = 5 \rightarrow b = 1.5 \) (not valid)
     - \( a = 6 \rightarrow b = -1 \) (not valid)
   - Valid four-digit palindromes: 2332, 3553.

5. **Five-digit Palindromes**: 
   - These have the form \( \overline{abcba} \). The sum of the digits is \( 2a + 2b + c = 13 \).
   - Solving for \( c \), we get \( c = 13 - 2a - 2b \). Both \( a \), \( b \), and \( c \) must be digits (1 through 9).
   - We need \( 13 - 2a - 2b \) to be a digit (1 through 9).
   - Possible values for \( a \) and \( b \) are 1 through 6. For each combination, we check if \( c \) is a valid digit:
     - \( a = 1 \rightarrow 2a = 2 \), so \( 11 - 2b \) must be a digit. This gives us:
       - \( b = 1 \rightarrow c = 9 \) (valid)
       - \( b = 2 \rightarrow c = 7 \) (valid)
       - \( b = 3 \rightarrow c = 5 \) (valid)
       - \( b = 4 \rightarrow c = 3 \) (valid)
       - \( b = 5 \rightarrow c = 1 \) (valid)
     - \( a = 2 \rightarrow 2a = 4 \), so \( 9 - 2b \) must be a digit. This gives us:
       - \( b = 1 \rightarrow c = 5 \) (valid)
       - \( b = 2 \rightarrow c = 3 \) (valid)
       - \( b = 3 \rightarrow c = 1 \) (valid)
     - \( a = 3 \rightarrow 2a = 6 \), so \( 7 - 2b \) must be a digit. This gives us:
       - \( b = 1 \rightarrow c = 3 \) (valid)
       - \( b = 2 \rightarrow c = 1 \) (valid)
     - \( a = 4 \rightarrow 2a = 8 \), so \( 5 - 2b \) must be a digit. This gives us:
       - \( b = 1 \rightarrow c = 1 \) (valid)
   - Valid five-digit palindromes: 19291, 27372, 35453, 43534, 51615, 62726, 73827, 81918, 93019.

6. **Six-digit Palindromes**: 
   - These have the form \( \overline{abccba} \). The sum of the digits is \( 2a + 2b + 2c = 13 \).
   - Solving for \( c