Fill in the blanks with a single digit such that the resulting sentence is true.
"In this sentence, the number of occurences of the digit 0 is ______, of the digit 1 is ______, of the digit 2 is ______, of the digit 3 is ______, of even numbers is ______, of odd numbers is ______, and of prime numbers is ______."
Previous iteration: #743084 (answer in #743754). Also, @Scroogey ended up solving the puzzle from 2 days ago: #741908
Brute force finds two solutions:
1 2 3 2 5 6 7 1 2 3 2 6 5 6
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I think it's an impossible problem to solve!
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This would be an impossible variant to this self-referential problem:
"In this sentence, the number of occurences of the digit 0 is ______, of the digit 1 is ______, of the digit 2 is ______, of even numbers is ______, of odd numbers is ______, and of prime numbers is ______."
(notice the omission of the digit 3 part)
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What's wrong with either of my solutions?
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In this sentence, the number of occurences of the digit 0 is 1, of the digit 1 is 2, of the digit 2 is 3, of the digit 3 is 2, of even numbers is 5, <---- of odd numbers is 6, and of prime numbers is 7.
In this sentence, the number of occurences of the digit 0 is 1, of the digit 1 is 2, of the digit 2 is 3, of the digit 3 is 2, of even numbers is 6, <---- of odd numbers is 5, and of prime numbers is 6.
I'm considering that 0 is neither even nor odd
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I think zero is generally considered even as it can be divided by two without remainder.
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Since the number zero is even, both are correct!
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Yes, there should be two valid solutions.
I'll post the reasoned solution to this problem at the same I post the next problem tomorrow.
I like brute force solving problems in my daytime job. It's sometimes so much faster.
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Reasoned solution:
Let the blanks be filled in with , , , , , , and , respectively. Among these, the first four entries satisfy for , while the other three entries each satisfy . Thus, , , and . Given , the values have opposite parity, so there’s at least one additional even and odd entry. Consequently, and . This limits to only five pairs: or . Each pair has exactly one prime number, hence , but , as this β€œ3” would add a fourth prime to the set, implying .
Now, only and are potential placements for β€œ1,” so . If , there are three primes, including one from , but not . Setting would then leave only three primes, and any other would be too large. Thus, . Assume ; then . To assess feasibility, suppose , which implies , creating a contradiction with . Thus, . If , leads to more than three primes, so .
With , counting primes from yields five primes. Setting results in six primes, while yields five primes, a contradiction. Therefore, , implying because would introduce three β€œ2”s. Since only allows a third 2, we conclude and .
Thus, we find six primes with or , resulting in two possible solutions: and .
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This is the answer ChatGPT gave.
"In this sentence, the number of occurrences of the digit 0 is 2, of the digit 1 is 1, of the digit 2 is 2, of the digit 3 is 1, of even numbers is 3, of odd numbers is 3, and of prime numbers is 3."
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Yeah, it's bad at this.
For other daily puzzles, i checked, it did arrive to the solution. But I'll just keep assuming people do it for the fun of solving rather than the few sats i give as rewards here and there. No way to police this anyhow :)
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I wonder why they haven't built in the capability for ChatGPT to simply say, "I don't know." The solution it gave is so obviously wrong.
(I haven't used it for previous problems, not that anyone accused me. But this one seemed especially suitable to test on chatgpt, and it failed pretty miserably.)
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β€œIn this sentence, the number of occurrences of the digit 0 is 1, of the digit 1 is 11, of the digit 2 is 2, of the digit 3 is 1, of even numbers is 2__, of odd numbers is 1, and of prime numbers is __2."
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  1. For the digit 0: Note the number of occurrences of "0" in the sentence after filling in the blanks. Example: # There are X occurrences of the digit 0.
  2. For the digit 1: Count the occurrences of "1" after filling in the blanks. Example: # There are X occurrences of the digit 1.
  3. For the digit 2: Count the occurrences of "2" after filling in the blanks. Example: # There are X occurrences of the digit 2.
  4. For the digit 3: Count the occurrences of "3" after filling in the blanks. Example: # There are X occurrences of the digit 3.
  5. For even digits: Count all even digits in the final sentence and confirm their total. Example: # There are X even digits.
  6. For odd digits: Count all odd digits in the sentence and verify their total. Example: # There are X odd digits.
  7. For prime numbers: Count the prime digits (2, 3, 5, 7) and confirm their total. Example: # There are X prime digits.
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23 minutes later ....
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sounds a bit like math gymnastics to me. :(
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Ain't it fun? :)
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In this sentence, the number of occurrences of the digit 0 is 3, of the digit 1 is 2, of the digit 2 is 2, of the digit 3 is 2, of even numbers is 5, of odd numbers is 4, and of prime numbers is 4.
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Once, once, once, once, once, twice, twice.
😁
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