Easy math puzzles
Poll
 | 21 votes (45.65%) | ||
| | 14 votes (30.43%) | ||
| | 6 votes (13.04%) | ||
| | 3 votes (6.52%) | ||
| | 12 votes (26.08%) | ||
| | 3 votes (6.52%) | ||
| | 6 votes (13.04%) | ||
| | 5 votes (10.86%) | ||
| | 12 votes (26.08%) | ||
| | 10 votes (21.73%) |
46 members have voted
May 21st, 2025 at 7:17:21 AM
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Ace2,
So would 22, 44, 66 qualify, since they have 11 in common? Your calculation seems to preclude this set, since they also share a common factor of 2.
Dog Hand
So would 22, 44, 66 qualify, since they have 11 in common? Your calculation seems to preclude this set, since they also share a common factor of 2.
Dog Hand
May 21st, 2025 at 11:10:26 AM
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Good catch. One must be very precise when using sets/logical arguments/inclusion-exclusion. I modified my answer below (in caps).Quote: DogHandAce2,
So would 22, 44, 66 qualify, since they have 11 in common? Your calculation seems to preclude this set, since they also share a common factor of 2.
Dog Hand
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Therefore, the chance that three random integers share a MINIMUM factor >10 is 0.1681 - 0.1666 = 0.15%.
Put another way, over 99% (1666/1681) of SETS WITH shared factors will INCLUDE 2,3,5 or 7.
It’s all about making that GTA
