Look at TX (the real solver)
{
  https://www.blockchain.com/explorer/transactions/btc/8c8ec6b3511c62500ea9b3a1c30ca937e15d251b55d30290a2a6da2f1124f3fb
}

It is safe to assume once he found the key, he accessed his wallet using electrum (or similar) and swept the #66 address (13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so)
{
  You can sweep a key with the WIF -> p2pkh:KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZfFoWMiwBt943V7CQeX
}

Now, that is a massive mistake, since there were 34 UXTO and only 2 actually worth taking.
{
 1) 0.066 BTC
 4) 5.940 BTC
}

About a minute later this TX shows up.
{
  https://www.blockchain.com/explorer/transactions/btc/57a88f47e4c047740b782a5562fca143ce85de0373cbff3a7d406e9ae7fc2f5f
}

Now, this one looks scripted.
{
 A) Only the highest UXTO was selected (5.940 BTC)

 B) The fee offered was 76,395 sats (+50K sats than first TX)
}

So miners dropped the first TX and replaced with the second TX, since it gave them more sats per byte.

The idea here is once you broadcast a TX, the pubkey shows up in the mempool.

Once the bot had the pubkey, it found the secret number in seconds.

And the problem with #66 is the secret number is very very small [46346217550346335726]

And this can possibly happen to keys all the way to #129, assuming the bot has the required precomputed DPs.

This is a theory, you have to do a lot of research on the topic if you want the real answer to #66.