Sorry to say, Claude Sonnet 4.5 (with and without extended thinking) has this one solved now. So, not “any” LLM, but maybe some still need work.
I presume they finally got around to training the internal reasoning loop into predicting when it needs to apply answer predictions character-by-character.
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Transcripts from Claude Sonnet 4.5, initially with extended thinking:
> How many a’s in the phrase “artichokes and aardvarks”? Be sure to check your answer.
> Let me count the letter ‘a’ in “artichokes and aardvarks”:
• artichokes: 1 ‘a’ (at the beginning)
• and: 1 ‘a’ (at the beginning)
• aardvarks: 3 ‘a’s (two at the beginning: “aa”, and one in the middle: “ardvarks”)
Total: 1 + 1 + 3 = 5 ‘a’s
Let me verify by marking them: artichokes and aardvarks
Yes, there are 5 letter ‘a’s in the phrase “artichokes and aardvarks”.
> Are you sure that’s correct?
> Let me recount more carefully:
“artichokes and aardvarks”
Going through each word letter by letter:
1. artichokes - has 1 ‘a’ (first letter)
2. and - has 1 ‘a’ (first letter)
3. aardvarks - has 3 ‘a’s (first letter, second letter, and one more in the middle)
Total: 1 + 1 + 3 = 5 ‘a’s
Yes, I’m confident the answer is 5. Is there something specific you’d like me to recheck?
> How many r’s in strawberry?
> Let me count the letter ‘r’ in “strawberry”:
s-t-r-a-w-b-e-r-r-y
Going through each letter:
• s: not an r
• t: not an r
• r: 1st r
• a: not an r
• w: not an r
• b: not an r
• e: not an r
• r: 2nd r
• r: 3rd r
• y: not an r
There are 3 r’s in “strawberry”.
> How many u’s in universalulis?
> Looking at “universalulis” letter by letter:
u-n-i-v-e-r-s-a-l-u-l-i-s
There are 2 u’s in “universalulis”.