I 've found an interesting table of numbers in an old issue of JMR, dedicated to astounding trivia regarding primes.
Erasing all the digits in the table's footnotes I got a challenging, albeit solvable puzzle:
The XX consecutive primes from X to XX sum up to the prime number XXX.
Also when arranged in groups of three, each group sums up to a prime number.
Furthermore, those partial sums with their digits reversed, also sum up to the same sum as before the reversal!
Try to reconstruct the trivia : both the table and the text.
list
10 repeat
20 P1=nxtprm(P1):P2=P1:Tot=P1:Ct=1
30 for I=1 to 8:P2=nxtprm(P2):Tot=Tot+P2:Ct=Ct+1:next
40 repeat
50 P2=nxtprm(P2):Ct=Ct+1
60 Tot=Tot+P2
70 if prmdiv(Tot)=Tot and Tot>99 and Tot<1000 and P2<100 and P1<9 the
n
80 :gosub *CheckIt
90 until P2>99
100 until P1>9
110 end
120 *CheckIt
130 ChkTot=0
140 Px=P1:ChkTot=P1:print Px;
150 repeat
160 Px=nxtprm(Px):print Px;
170 if Px<10 then ChkTot=ChkTot+Px
180 if Px>10 then ChkTot=ChkTot+10*(Px@10)+Px\10
190 until Px=P2:print
200 if nxtprm(ChkTot)=ChkTot then print P1;P2,Tot,ChkTot
205 print Ct,P1;P2,Tot,ChkTot:print
210 return
OK
The UBASIC program above finds sequences that start with a single digit prime and extend at least 10 primes to a two-digit prime, and add up to a prime.
I've marked off part of its output based on criteria shown below:
run
2 3 5 7 11 13 17 19 23 29 31 37
12 2 37 197 431
2 3 5 7 11 13 17 19 23 29 31 37 41 43
14 2 43 281 479
3 5 7 11 13 17 19 23 29 31 37 41 43 47 53
15 3 53 379 586
3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61
17 3 61 499 697
5 7 11 13 17 19 23 29 31 37 41
11 5 41 233 440
5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67
17 5 67 563 770
7 11 13 17 19 23 29 31 37 41 43
11 7 43 271 469
-----------------------------------------------------------
7 11 13 17 19 23 29 31 37 41 43 47 53 59 61
15 7 61 491 689
-----------------------------------------------------------
7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83
89
21 7 89 953 1052
Only multiples of 3 in the count of primes were further tested via spreadsheet, which included the 15-member series marked off above.
2 10
3 15 |371| 13
5 23 |131| 95
7 31 | 79| 79
11 41 | 95| 131
13 49 | 13| 371
17 59 *689* 112
19 71 152
23 83 953
29 97
31 109 38
37 121 94
41 131 32
43 143 164
47 159
53 173
59 187
61 199
67 211
71 223
73 235
79 251
83
89
The columns on the spreadsheet show the primes from 2 to 89 and the triplet sums starting at any of the three possible starting points.
As the 15-member sequence does indeed have a multiple of three members, the series of triple totals that starts at 7, starts the totals with 31 (=7+11+13) and includes 59, 97, 131 and 173. The basic program had already calculated the total of the reversed digits primes themselves as 689. Off to the side, on the spreadsheet, the five triplet totals, with digits reversed, also add to 689. I think this is what is sought, but then again what it matches is not the sum of the non-reversed groups of three but rather the reversed-digit individual primes of the 15-prime series.
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Posted by Charlie
on 2010-05-20 22:44:12 |
Is this it?
