Replace the numerals 1 through 8 with ever increasing prime numbers, always using the next lowest possible that is available
1 to fulfill the criteria on the left. Then do the same for the right.
Present a series for the left, and one for the right.
7 + 8 = Square
6 = Prime
4 + 5 = Square
1 + 2 + 3 = Square
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7 + 8 = Cube
6 = Prime
4 + 5 = Cube
1 + 2 + 3 = Cube
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If it was required that the Right set required "1" to be the next Prime following on after that used for the "8" in the Left set, what might the Right set read, if indeed it is possible?
1.
Note,
"always using the next lowest possible that is available" means that if it is next on the list it cannot be dismissed unless it is
the last of a group of two or three and will not fulfill the criterion. Only then may you advance to the next.
Part 1a:
list
10 dim Pr(16)
15 input Strt
20 Pr(1)=Strt
30 gosub *Addon(2)
40 end
200 *Addon(N)
210 X=nxtprm(Pr(N-1))
220 if N=3 and fnSq(Pr(1)+Pr(2)+X)=0 then X=nxtprm(X):goto 220
230 if N=5 and fnSq(Pr(4)+X)=0 then X=nxtprm(X):goto 220
240 if N=8 and fnSq(Pr(7)+X)=0 then X=nxtprm(X):goto 220
250 Pr(N)=X
260 if N=8 then
270 :for I=1 to 8:print Pr(I);:next:print:end
280 :else
290 :gosub *Addon(N+1)
300 return
500 fnSq(N)
510 Sr=int(sqrt(N)+0.5)
520 if Sr*Sr=N then return(1):else return(0)
530 return
OK
In the below runs, the "No gosub" response at the end has been omitted (resulting from ending the program with gosubs (calls) pending).
run
? 2
2 3 11 13 23 29 31 113
run
? 3
3 5 17 19 557 563 569 587
run
? 5
5 7 13 17 19 23 29 71
run
? 7
7 11 31 37 107 109 113 211
run
? 11
11 13 97 101 223 227 229 347
run
? 13
13 17 19 23 41 43 47 53
run
? 17
Break in 220
run
? 19
19 23 79 83 113 127 131 193
run
? 23
23 29 173 179 397 401 409 491
Note that a satisfactory solution starting with 17 was not being found, so the program was halted.
Part 1b:
list
10 dim Pr(16)
15 input Strt
20 Pr(1)=Strt
30 gosub *Addon(2)
40 end
200 *Addon(N)
210 X=nxtprm(Pr(N-1))
220 if N=3 and fnCu(Pr(1)+Pr(2)+X)=0 then X=nxtprm(X):goto 220
230 if N=5 and fnCu(Pr(4)+X)=0 then X=nxtprm(X):goto 220
240 if N=8 and fnCu(Pr(7)+X)=0 then X=nxtprm(X):goto 220
250 Pr(N)=X
260 if N=8 then
270 :for I=1 to 8:print Pr(I);:next:print:end
280 :else
290 :gosub *Addon(N+1)
300 return
500 fnCu(N)
510 Cr=int((N)^(1/3)+0.5)
520 if Cr*Cr*Cr=N then return(1):else return(0)
530 return
run
? 2
2 3 59 61 1667 1669 1693 4139
run
? 3
3 5 19 23 41 43 47 953
run
? 5
5 7 113 127 1601 1607 1609 25391
run
? 7
7 11 107 109 1619 1621 1627 6373
run
? 11
11 13 101 103 113 127 131 1597
run
? 13
13 17 313 317 683 691 701 38603
run
? 17
17 19 89 97 2647 2657 2659 30109
run
? 19
19 23 83 89 127 131 137 863
run
? 23
23 29 73 79 137 139 149 1579
run
? 29
29 31 283 293 3803 3821 3823 4177
No gosub
Part 2:
Changing line 20 of the cube version to
20 Pr(1)=nxtprm(Strt)
Starting with the terminal numbers from part 1a:
run
? 113
127 131 11909 11923 15077 15083 15091 58997
run
? 587
593 599 8069 8081 18919 18947 18959 27697
run
? 71
73 79 191 193 2551 2557 2579 3253
run
? 211
223 227 881 883 1861 1867 1871 11953
run
? 347
349 353 4211 4217 22783 22787 22807 152809
run
? 53
59 61 223 227 773 787 797 3299
run
? 193
197 199 1801 1811 4021 4027 4049 6599
run
? 491
499 503 3911 3917 9907 9923 9929 87407
No gosub
The best in terms of the lowest final number seems to be:
29 71 sq=100
23
17 19 sq=36
13 sq=25
7
5
followed by
2579 3253 cu=5832
2557
193 2551 cu=2744
191 cu=343
79
73
and also, for low numbers in general
47 53 sq=100
43
23 41 sq=64
19 sq=49
17
13
followed by
797 3299 cu=4096
787
227 773 cu=1000
223 cu=343
61
59
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Posted by Charlie
on 2010-01-06 15:44:13 |
computer solution
