| ▲ | 8bitsrule 12 hours ago | ||||||||||||||||
A few months ago I asked a model how many primes are divisible by 35 with a remainder of 6. It confidently replied 'none'. Counterexample: 35 + 6. | |||||||||||||||||
| ▲ | buzzin__ 12 hours ago | parent | next [-] | ||||||||||||||||
Kimi 2.6 gives the answer ""By Dirichlet's theorem on arithmetic progressions, since gcd(6,35)=1 , there are infinitely many primes of the form 35k+6 . So the answer is: infinitely many primes give a remainder of 6 when divided by 35. | |||||||||||||||||
| ▲ | buzzin__ 12 hours ago | parent | prev | next [-] | ||||||||||||||||
But, if the reminder is 6, they are not really divisible, are they? Try again with a sentence that actually makes sense: "How many primes, when divided by 35, give a reminder of 6?" | |||||||||||||||||
| ▲ | jibal 12 hours ago | parent | prev [-] | ||||||||||||||||
non sequitur | |||||||||||||||||
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