No.
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Two logicians achieve infinite intelligence and they sit around letting random observers scribble numbers on their foreheads?
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And then don’t tell each other the number they can see.
“Yours is 25.”
“So’s yours. Wanna get a coffee?”
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Infinitely intelligent does not equate to having any ‘common sense’ 
(And @Kilkrazy and @anon33176345 )
I can only figure that this is some kind of Saw scenario
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If Alice’s number is 5 and she sees a 10, she doesn’t know - she just knows she could have a 5 or a 40.
But in that situation, Bob then either sees a 40 (in which case he knows he has a 10) or sees a 5 (in which case he knows Alice did not see a 45, so knows he has a 10).
If you follow that track you will find the answer. There’s only one number for Alice to have where Bob can not make such a deduction.
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Killing time until the foursome with the one guard who always tells the truth and the other one who always lies. (“I’m not coming! I’m not coming!”)
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When Alice says “I don’t know” - then Bob would know his is 10 - because Alice would know if she saw 45? Too bad Bob can’t use Alice’s uncertainty to change his answer.
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By the time Bob says he doesn’t know, then Alice MUST know her number.
If she didn’t know, then WE couldn’t deduce it either. Because by then she knows everything we know. PLUS she knows Bob’s number.
As others have determined, Alice’s number is 25 and Bob’s is 2.
When Alice sees Bob has 2, then she knows hers must be either 25 or 48. When Bob says he doesn’t know, then she and we know she has 25.
Because if Bob saw 48, then he would have known his number can only be 2. But he didn’t.
We know it for more complex reasons than Alice, because we don’t have the advantage she has of knowing Bob’s number at the start. We have to determine there is only ONE combination that allows the result to be known only AFTER the second person says they can’t deduce the answer.
Bob’s reasoning is more complex than Alice’s first is. Because at first she only knows Bob’s number. But Bob knows her number PLUS that his doesn’t allow her to deduce hers from his alone.
Amusingly, it took ChatGPT-4o two tries to get this right. Even though it said this is a well-known logic puzzle.
First try it gave an indeterminate answer claiming both could have either of two values. I just responded “Wrong” and it found the correct answer on the second try.
At least it didn’t cheat. 
[ETA] I (and ChatGPT) was still wrong here. I thought we could know Bob’s number too. But we don’t.
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Bob never knows his number. Neither do we. His could be either 25 or 2. Alice figuring out that her number is 25 doesn’t change that. The only person who knows what Bob has is Alice.
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Me? No. But Chat GPT has it figgured:
a riddle for you: Alice and Bob are two infinitely intelligent logicians. Each has a number drawn on their forehead. Each can see the other’s number but not their own. Each knows that both numbers are positive integers. An observer tells them that the number 50 is either the sum or the product of the two numbers. Alice says to Bob, “I do not know my number,” and Bob replies, “I do not know my number either.” What is Alice’s number?
ChatGPT
Let’s break down the riddle step by step to determine Alice’s number.
Given:
Summary
- Alice and Bob are both infinitely intelligent logicians.
- Each has a number on their forehead, and they can see each other’s number but not their own.
- The observer tells them that 50 is either the sum or the product of their numbers.
- Alice says, “I do not know my number.”
- Bob replies, “I do not know my number either.”
Analysis:
- Possible pairs of numbers:
- If 50 is the sum of Alice’s number (A) and Bob’s number (B), then possible pairs (A,B)(A, B)(A,B) are: (1,49)(1, 49)(1,49), (2,48)(2, 48)(2,48), (3,47)(3, 47)(3,47), (4,46)(4, 46)(4,46), …, (24,26)(24, 26)(24,26), (25,25)(25, 25)(25,25).
- If 50 is the product of Alice’s number (A) and Bob’s number (B), then possible pairs (A,B)(A, B)(A,B) are: (1,50)(1, 50)(1,50), (2,25)(2, 25)(2,25), (5,10)(5, 10)(5,10).
- Alice says, “I do not know my number.”
- If Alice had seen 1 on Bob’s forehead, she would have known her number must be either 49 (sum case) or 50 (product case), and she would know immediately because 50 cannot be the sum. Thus, Alice does not see 1 on Bob’s forehead.
- If Alice had seen 50 on Bob’s forehead, she would know her number must be 1 (product case). Therefore, Alice does not see 50 on Bob’s forehead.Alice’s statement implies that Bob’s number is neither 1 nor 50.
- Bob says, “I do not know my number either.”
- Bob also cannot have seen 1 or 50 on Alice’s forehead for the same reasons.
- If Bob had seen 49 on Alice’s forehead, he would know his number must be 1 (sum case). Therefore, Bob does not see 49 on Alice’s forehead.
- If Bob had seen 25 on Alice’s forehead, he would know his number must be 2 (product case). Therefore, Bob does not see 25 on Alice’s forehead.Bob’s statement implies that Alice’s number is neither 1, 25, nor 49.
Conclusion:
- The only pair left that satisfies all the conditions (both Alice and Bob not knowing their numbers) is A=2A = 2A=2 and B=25B = 25B=25, where the sum is 272727 and the product is 505050.
- However, since we ruled out 25 based on Bob’s statement, let’s reconsider the remaining pairs.
Since none of the other pairs work under the same logic, we realize there must have been a misstep. Rechecking our pair possibilities, we find that the correct pair is:
- A=5A = 5A=5, B=10B = 10B=10 (sum is 15, product is 50).
Given this pair, if Alice sees 10 on Bob’s forehead, she would not immediately know her number, because 10+A10 + A10+A does not sum to 50, and A×10=50A \times 10 = 50A×10=50 is possible. Bob, seeing 5, would not know either since 5+B5 + B5+B and 5×B5 \times B5×B could correspond to several pairs. Thus, Alice’s number is 5.
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Yeah, about the normal intelligence level of AI, giving the wrong answer with confidence
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That whole explanation is riddled with errors, and the result is wrong.
Reading it is - as I can sadly confirm - a complete waste of time.
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Nonsense.
Each knows that both numbers are positive integers. These “answers” ain’t.
Alice sees 49, and lies in appreciation for deadpanning.
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OMG, this should have been oblig from the start!!! 
x 100
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Alice’s number is 25. Bob’s may be either 2 or 25.
Once Bob says he doesn’t know his, then Alice (and we) know her number is 25. But only Alice can possibly know Bob’s number, by seeing it. Either Alice has 25 and Bob has 2 or Alice has 25 and Bob has 25. Those are the only 2 combinations that end with Bob being unable to deduce his number after Alice couldn’t deduce hers until she knew Bob couldn’t deduce his.
All other combinations allow Alice to deduce her number from the start, or allow Bob to deduce his either from Alice’s number alone, or in combination with knowing Alice could not deduce hers from his.
If Alice sees Bob has 2, then she can only know her number must be 25 or 48. But if Bob saw 48, then he would know his number could only be 2. Since he doesn’t know, then hers can only be 25.
I was almost there, but didn’t quite recognize it because my process was confusing. The Futility Closet post explains it as concisely as possible. By a more rigorous, but not obvious (to me) approach. Only after reading the article did I realize I had the answer but hadn’t noticed. Because I was thinking we would know BOTH numbers. We can’t.
[ETA] The gnarly thing about this is that, in the end, everyone knows Alice’s number, but only Alice can possibly know Bob’s number.
I kept confusing myself by thinking “If we can know the answer, then how can Bob and Alice not also know the answer?” But they can and do know “the answer”. I wasn’t considering that the answer to the puzzle is different from the answer to the question Bob and Alice had to answer.
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