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Topic: Prime pairs (Read 947 times) |
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BNC
Uberpuzzler
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Prime pairs
« on: Mar 25th, 2003, 3:09pm » |
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1. Can you arrange the natural numbers 1..10 into 5 pairs, such that the sums of the pairs will be 5 different prime numbers? 2. Repeat 1 for 1..50 (25 pairs, 25 different prime numbers)
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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Icarus
wu::riddles Moderator Uberpuzzler
Boldly going where even angels fear to tread.
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Re: Prime pairs
« Reply #1 on: Mar 25th, 2003, 3:36pm » |
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For (1) there is :1+4, 2+5, 3+8, 6+7, 9+10
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"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? " - Anonymous
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NickH
Senior Riddler
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Re: Prime pairs
« Reply #2 on: Mar 25th, 2003, 3:39pm » |
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1. 2+3=5, 1+6=7, 4+7=11, 5+8=13, 9+10=19. 2. I'll have to come back to this one later...
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Nick's Mathematical Puzzles
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cho
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For no. 2: No. There are only 24 primes from 3 to 99.
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cho
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For a slightly harder question, what is the highest number for which you could solve the riddle? hint If I'm right, the solution for that number is quite simple at this point. Proving there are no higher solutions would be more difficult.
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mistysakura
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Re: Prime pairs
« Reply #5 on: Mar 27th, 2003, 2:30am » |
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on Mar 25th, 2003, 4:53pm, cho wrote:For no. 2: No. There are only 24 primes from 3 to 99. |
| Yes, but your sums can be higher than 99.
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BNC
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Re: Prime pairs
« Reply #6 on: Mar 27th, 2003, 3:56am » |
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on Mar 27th, 2003, 2:30am, mistysakura wrote: Yes, but your sums can be higher than 99. |
| No, they can't. 49+50 is the highest possible pair sum.
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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cho
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For the highest possible solution, the solution is simple. Take the 10 answer and add 11+12=23. I suggest that there is no higher answer. 14: The pairs sum to 105 but the available primes only add up to 98. (Primes must be higher than the pairs because some will not be used). 16: Pairs=136; Primes=158, but you can't use both 31 and 29 in the answer (16+15=31, 14+13 is only 27). 18: Pairs=171; Primes=158. No new primes were added, so primes fall short again. At this point you see you don't have to try all possibilities, just those that add new primes. 20: Pairs=210; Primes=195 22: Pairs=253; Primes=279, but you can't use both 43 and 41. 24: Pairs=300; Primes=326, but you can't use 47,43, and 41 all together. See, the highest numbers in the solution must average at least 4 apart. (24+23=47,22+21=43,20+19=39). You can not use primes that are 2 apart unless you've left one of these higher possibilities unused. From this point on the pairs total gradually outpaces the primes, and the only time the primes do some catching up is when you have consecutive primes, but then you can't use them both anyway. By the time you reach 50, the pairs are over 200 points ahead of the primes, and since primes become more and more widely separated, I don't believe they could ever catch up.
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