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Topic: nonagonal numbers (Read 1536 times) |
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LeoYard
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nonagonal numbers
« on: Feb 14th, 2011, 10:13am » |
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I found a puzzle about nonagonal numbers on another forum. The puzzle was unanswered, I'm confused about it. So I'm sharing it with this community.
A nonagonal number is a number of the form
n*(7n-5)/2.
The puzzle asks to find a nonagonal number that is the sum of three 3-digit prime numbers which together contain the nine nonzero digits once each.
What does it mean "contain the nine nonzero digits once each?
does it mean that these numbers are not allowed to be a 0 in there...and the 3 primes must contain all digits, if you scramble them together?
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towr
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Re: nonagonal numbers
« Reply #1 on: Feb 14th, 2011, 11:10am » |
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I think you're supposed to use the numbers 1..9, each once, to form 3 primes (with three digits each), that equal n*(7n-5)/2 for some (integer) n.
n will be some number smaller than or equal to 27 and greater than or equal to 16. (Because of the greatest and smallest possible sum you can make from three three-digit numbers that use each digit 1..9 once)
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LeoYard
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Re: nonagonal numbers
« Reply #2 on: Feb 14th, 2011, 12:09pm » |
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Thanks for clarifying the puzzle. I understand it better now.
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ThudnBlunder
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Re: nonagonal numbers
« Reply #3 on: Feb 14th, 2011, 2:32pm » |
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And I don't think it is cheating to use this link.
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| « Last Edit: Feb 15th, 2011, 3:30am by ThudnBlunder » |
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rmsgrey
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Re: nonagonal numbers
« Reply #4 on: Feb 15th, 2011, 5:09am » |
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on Feb 14th, 2011, 2:32pm, ThudnBlunder wrote:| And I don't think it is cheating to use this link. |
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So between you and towr we get this list of candidate nonagonal numbers:
856, 969, 1089, 1216, 1350, 1491, 1639, 1794, 1956, 2125, 2301, 2484
Finding a list of 3-digit primes without 0's or repeated digits is left as an exercise for the reader 
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JohanC
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Re: nonagonal numbers
« Reply #5 on: Feb 15th, 2011, 7:17am » |
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The funny thing is, only the nonagonal number is asked, not the 3 primes. In fact, it is quite straightforward from here on:
Primes of 3 digits are all odd, so the sum of 3 of them must be odd.
Given the digits of the 3 primes, we can calculate their sum modulo 9. There seems to be only one nonagonal number in the given range to fit this requirement.
Finding the 3 primes themselves doesn't seem to be too straighforward. It's easy to decide the last digit of each of them, but to arrange the first 2 digits of each of them, lots of combinations need to be checked.
As stated before, luckily the primes themselves are not asked. Maybe there even are multiple solutions?
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rmsgrey
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Re: nonagonal numbers
« Reply #6 on: Feb 16th, 2011, 11:57am » |
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on Feb 15th, 2011, 7:17am, JohanC wrote:The funny thing is, only the nonagonal number is asked, not the 3 primes. In fact, it is quite straightforward from here on:
Primes of 3 digits are all odd, so the sum of 3 of them must be odd.
Given the digits of the 3 primes, we can calculate their sum modulo 9. There seems to be only one nonagonal number in the given range to fit this requirement.
Finding the 3 primes themselves doesn't seem to be too straighforward. It's easy to decide the last digit of each of them, but to arrange the first 2 digits of each of them, lots of combinations need to be checked.
As stated before, luckily the primes themselves are not asked. Maybe there even are multiple solutions? |
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So you can show that, if any nonagonal number meets the conditions, it must be that one. What you haven't shown is that there is at least one that does...
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towr
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Re: nonagonal numbers
« Reply #7 on: Feb 17th, 2011, 1:38pm » |
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Given that we know the target nonagonal number; the primes have to end with an odd digit, and we know the last digits have to sum to 9 or 19; we can exclude 5, so the last digits are 3,7,9
We can take the list of 3-digit primes from http://primes.utm.edu/lists/small/1000.txt and start eliminating everything with a 0, ending in 1, having a 3,7, or 9 somewhere other than the end, and anything with double digits.
We can eliminate further primes by noting that the minimum sum of two 3-digit numbers with different digits is 381; this puts a limit on the maximum for the last prime.
We now know the middle 3 digits include the 8 and the other two sum to 9. So they're 4 and 5, leaving 1,2 and 6 for the first digits.
This leaves 149 as the smallest prime. And consequently the largest primes then has to end in 3, leaving 257 as the second prime. And so finally the last prime is 683. And together they add to 1089 as given.
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| « Last Edit: Feb 17th, 2011, 1:38pm by towr » |
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Benny
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Re: nonagonal numbers
« Reply #8 on: Feb 17th, 2011, 7:09pm » |
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I got : the only solution is 149+257+683=1089
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