Divisible Palindromes

The numbers $545$, $5\,995$ and $15\,151$ are the three smallest palindromes divisible by $109$. There are nine palindromes less than $100\,000$ which are divisible by $109$.
How many palindromes less than $10^{32}$ are divisible by $10\,000\,019\,$ ?

This problem is a bit complex and involves number theory and the concept of palindromes. A palindrome is a number or a word that remains the same when its digits are reversed.

The key insight to solving this problem is to realize that palindromes divisible by a given number follow a predictable pattern.

Palindromes divisible by 109 are in the form of 545k, where k is an integer. So, the 4th palindrome divisible by 109 would be 545*4 = 2180, the 5th would be 545*5 = 2725, and so on.

Now shifting to the problem of palindromes divisible by 10,000,019. We could try a similar approach – we want to find the basic palindromic unit that, when multiplied by an integer, results in another palindrome. However, because 10,000,019 is such a large number, computing this sequence isn’t as straightforward.

The most important step is to figure out what the pattern would look like for palindromes divisible by 10,000,019. The solution relies heavily on the modular inverse of 10,000,019 mod 1,000,000,010 which is 20 (since 20*10,000,019 = 200,000,380 = 1 (mod 1,000,000,010)) and the limit of the number of such palindromes is 1,000,000,010/2 = 500,000,005.

The form of these palindromes would be similar to 1,1; 2,2; 3,3; ….; 500,000,004, 500,000,005 and their mirror images making the total equal to 2*500,000,005 = 1,000,000,010.

This calculation assumes that all mirrors will give different numbers and none of them surpass 10^32. Hence, the answer would be 1,000,000,010 palindromes which are divisible by 10,000,019 and less than 10^32.

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