Let us consider a duodecimal prime number which when re-interpreted as a base ten positive integer is a composite number. As an example, (45)12 is equal to (53)10 and therefore a prime number. However, (45)10 is evidently a composite number.
Determine the probability an integer chosen at random from (100)12 to (BBB)12 inclusively is a prime number in the duodecimal system, but is composite when re-interpreted as a base ten number.
For v = fromBase("100", 12) To fromBase("bbb", 12)
overct = overct + 1
If prmdiv(v) = v Then
r$ = base$(v, 12)
dv = Val(r$)
good = 1
For i = 1 To Len(r$)
If InStr("0123456789", Mid(r$, i, 1)) = 0 Then good = 0: Exit For
Next
If good = 1 And prmdiv(dv) < dv Then
Text1.Text = Text1.Text & r$ & Str(v) & crlf
ct = ct + 1
End If
End If
Next v
g = gcd(ct, overct)
Text1.Text = Text1.Text & crlf & ct & Str(overct) & Str(ct / g) & Str(overct / g) & mform(ct / overct, "##0.000000000") & crlf
finds that 90 out of the 1584 cases fit the criterion, for a probability of 5/88 ~= 0.056818182.
base-12 prime base-12 number
base-10 composite translated to decimal
for verification of primality
105 149
111 157
117 163
125 173
141 193
145 197
147 199
171 229
175 233
195 257
205 293
217 307
221 313
225 317
237 331
255 353
267 367
285 389
291 397
295 401
301 433
315 449
321 457
325 461
327 463
357 499
365 509
375 521
377 523
391 541
415 593
427 607
435 617
437 619
447 631
455 641
465 653
471 661
481 673
485 677
497 691
507 727
511 733
517 739
527 751
531 757
535 761
545 773
565 797
575 809
585 821
591 829
611 877
615 881
637 907
655 929
665 941
675 953
687 967
695 977
705 1013
711 1021
721 1033
735 1049
737 1051
745 1061
747 1063
767 1087
771 1093
775 1097
785 1109
791 1117
801 1153
817 1171
825 1181
835 1193
841 1201
851 1213
855 1217
865 1229
867 1231
871 1237
901 1297
905 1301
921 1321
927 1327
955 1361
965 1373
987 1399
995 1409
stats:
90 1584 5 88 0.056818182
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Posted by Charlie
on 2014-12-26 14:58:12 |
computer solution
