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#1 |
May 2004
New York City
23×232 Posts |
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Construct a 5 x 5 square containing distinct primes
such that each row, column and diagonal sums to a distinct prime. |
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#2 |
1976 Toyota Corona years forever!
"Wayne"
Nov 2006
Saskatchewan, Canada
23·569 Posts |
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#3 |
May 2004
New York City
23×232 Posts |
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A magic square of primes (where every sum is the same) is
solved elsewhere (although it would be a perfectly good puzzle to re-solve). Here every sum is a different prime. |
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#4 |
May 2004
New York City
108816 Posts |
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The original problem was perhaps too computationally simple
to be interesting. The following additional condition adds an iota of complexity: The 25 distinct primes in the square should be the first 25 odd primes {3,5,7,...,97,101}. (I have a solution which wasn't hard to find by trial and error, so there must be many solutions; but plain brute force on the 25! such possible squares is obviously too computationally costly.) Last fiddled with by davar55 on 2008-05-13 at 18:08 |
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#5 |
Oct 2007
Manchester, UK
24748 Posts |
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Here's one:
Code:
239
/
/
/
3, 5, 7, 11, 17 --- 43
13, 19, 29, 23, 43 --- 127
31, 67, 61, 47, 71 --- 277
53, 59, 41, 73, 37 --- 263
79, 83, 89, 97, 101 --- 443
| | | | | \
| | | | | \
| | | | | \
179 233 227 251 269 257
Last fiddled with by lavalamp on 2008-05-17 at 02:26 |
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#6 |
Sep 2006
Brussels, Belgium
13·127 Posts |
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There is an error in your calculations : the last row total is of by 6. But the right number is prime so the solution stands :-)
Jacob |
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#7 |
Oct 2007
Manchester, UK
22·5·67 Posts |
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Hm, I think I worked the total out right, but wrote it down wrong.
It wasn't just a fluke, honest! ;) |
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#8 |
Feb 2007
24×33 Posts |
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Does the sequence
a(n) = number of square matrices containing the first (2n+1)x(2n+1) odd primes, such that row, column and diagonal sums are distinct primes exist on OEIS ? |
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#9 |
Oct 2007
Manchester, UK
22×5×67 Posts |
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It would appear that there are an awful lot of these out there, so perhaps the challange should be to find a square with the lowest standard deviation of column/row/diagonal totals.
I'll start the ball rolling with a slightly modified version of the last square I posted, with an s.d. of 84.51: Code:
239
/
/
/
3, 5, 7, 11, 17 --- 43
13, 19, 29, 23, 43 --- 127
31, 67, 61, 47, 71 --- 277
53, 59, 89, 73, 37 --- 311
79, 83, 41, 97, 101 --- 401
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179 233 227 251 269 257
Last fiddled with by lavalamp on 2008-05-19 at 15:06 |
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#10 |
May 2004
New York City
23·232 Posts |
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Here's another solution:
041 005 007 071 003 013 023 029 031 067 059 053 043 047 037 019 011 089 061 017 079 101 083 097 073 Rows: 127,163,239,197,433 Columns: 211,193,251,307,197 Diagonals: 241,167 (Standard deviation: 76.7) An alternative measure is simply mini-max: minimize the largest sum. By that measure, lavalamp's solution is a better one. |
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