Problems & Puzzles:
Puzzles
Puzzle
579.-
A triangular array with primes
Jean Brette sent the following nice
puzzle:
A triangle is divided in n^2 small
triangles, and colored like a chessboard. (here n = 4)
Fill it with distinct primes with the
following rule : A prime in a white triangle is equal to the sum of his
three neighbors, in the grey triangles. Examples (S is the sum of the
primes used)
Q1. Could you find
similar triangles for n > 5 ? (with the smallest S, if possible)
Q2. Does it exist a solution for each n ?
Jean Brette himself found one solution
using consecutive primes! here is it:
n=3, S=127, Consecutive
primes
17
7 29 5
13 23 3 19
11
Q3. Does it exist
solutions with consecutive primes for n>3?
Q4. Does it exist
solutions using consecutive integers?
Contributions came from Jean Brette (BTW, all of his
solutions came without using a PC), Hakan Summakoğlu & Robert D. Mohr.
***
Jean Brette wrote:
Q3. n=4, S= 438, Consecutive primes!!!
5
13 41
23
7 37 17 43 3
29 47 11
59 31 53
19
Q4.
There is no solution with consecutive and positive integers starting
with 1- You find the proof attached.
But if you can use zero or negative numbers:
1) If we can use the zero, here are solutions with n = 2 and n = 3
2
0 3 1
--------------------------
2
0 5 3
6 7 1 8 4
2) if we can use negative numbers, here are solutions with n = 2 and n
= 3
2
0 1 -1
------------------------
With -2, -1, 0, 1, 2, 3, 4, 5, 6
3
4 5 -2
2 6 0 -1 1
With -3, - 2, -1, 0, 1, 2, 3, 4, 5
-2
4 5 3
-3 1 0 2 -1
***
Hakan wrote:
Q1:
|
n=6,S=3116 |
|
|
3 |
|
|
5 19 11 |
|
|
7 29 17 41 13 |
|
|
23 61 31 101 53 103 37 |
|
|
43 113 47 137 59 179 67 193 89 |
|
|
109 223 71 191 73 211 79 229 83 269
97 |
|
|
|
|
n=7,S=6283 |
3
|
|
5 19 11 |
|
7 29 17 41 13 |
|
23 61 31 101 53 103 37 |
|
43 113 47 137 59 179 67 193 89 |
|
71 197 83 227 97 229 73 271 131 359
139 |
|
157 379 151 313 79 283 107 307 127 367
109 397 149 |
Q3:
n=5, S=1159, Consecutive primes
|
3
|
5 61 53
|
29 71 37 101 11
|
31 83 23 73 13 67 43
|
41 89 17 59 19 79 47 97 7
|
***
Robert wrote:
For Q4, the n=3 case a solution with
lowest possible S value is S=48, an "almost-consecutive":
6
1
9
2
4
8
3
10
5
***
Robert D. Mohr wrote on March 2011:
Seven solutions for n=4 using consecutive primes:
5
5
"41" "41"
23 13 7
29
"43" "37"
"31" "43"
"53" "59"
"47" "53"
"47" "59"
11
13
"59" "47"
17
31 3
31
"29"
"41" "29" "43"
"47" "43" "53" "59" "41"
"53"
11 11
"41"
"47"
17
13
5 31
"59"
"43" "29"
"41"
"53' "31" "47" "59" "43"
"53"
23
"41"
11 7
"47" "29"
17 19 3
"53" "37" "59"
31 5 13 43
***
Jean Brette added on March 11:
A) Congratulations
to Hakan and his solution for Q3 and n = 5 !!!
If he has used a computer, it will be interesting to use it again to find
ALL the solutions for Q3 with n= 4
My feeling tells me that they are many.
B) Q1 ; n = 6
My best solution is :
23
127
229 79
3 173 43
139 17
71 103 29
109 37 61 7
13 89 5 101 67
151 47 107 53
41 113 59
83 19 97 11
131 73 157 31
( among the 36 first odd primes, 149 and 137
was not used, and was replaced by 229 and 173)
So S = 2582
+ (229 + 173) – (149+137)= Sp
+ 116 = 2698 where
Sp = 2682 is the sum of the 36 first odd primes.
C) Q4 ; n = 3.
a) The minimal solution given by Robert Mohr is quite surprising, since it
uses “almost consecitive and positive numbers”
So we have two more questions :
“Question 5” : What are the minimal solutions for n>3 ? (with positive
integers)
“Question 6” : The numbers in these solutions are they “almost
consecutive” ?
b) An answer to Q4 and Q6;
n =3
There are two , and only two, distinct solutions for n=3 (proof
attached Pz 579 Q6.doc )
5
1 8 2
6 10 3 9 4
and the Robert’s solution.
c) Question 5, n = 4
I’ve found three solutions with sum S= 142. The two first
use the integers (1,2,.....13, 14, 17, 20)
The third is very nice, with only one gap (14, 15) and (17, 18) added. So
the integers are : ( 1, 2, .....12, 13, 16, 17, 18)
The ten first are grey.
8
7 18 3
2 13 4
12 5
9 17 6 11 1
16 10
Since they are “hand made”, I don’t know if they are minimal. :o(
D) Filling the area with positive odd
numbers. (n = 3)
With the same questions (Q5 and Q6)
Curiosly, this time, there are solutions for n = 3 with consecutive
odd numbers.
7 11
5 13 1
5 17 1
9 17 3 15 11
and 7 15 3 13 9
***
Robert D. Mohr added:
OK this took about 36 hours to run. I'll
have to improve the elegance of my code for n=6 and higher or it will
take quite a bit longer.
Anyway, believe it or not there are 2949
solutions using consecutive primes for n=5 with S=1159.
I have enclosed a
notepad document with all solutions. Note that each solution is
given as a triangle with the 15 gray squares first with a second
triangle showing the 10 white squares immediately below.
***
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