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 Self-Cardinality (Posted on 2010-05-07)
Arrange nine digits in a 3x3 array such that each different digit that appears appears as many times as its cardinality, so for example, if the digit 3 appears at all, it appears 3 times. Also the eight 3-digit numbers formed from the rows, columns (top to bottom) and two diagonals (also top to bottom) must all be different.

Also, the sum of the nine digits used cannot itself be written using the digits you've included in the array.

What is that sum of these nine digits?

 Submitted by Charlie Rating: 3.0000 (1 votes) Solution: (Hide) The different digits used must sum to 9, so that all the positions in the array are filled and no more. The sets of different digits can be: ``` 9 8 1 7 2 6 3 6 2 1 5 4 5 3 1 4 3 2 ``` Clearly, nine 9's will not allow for all eight 3-digit numbers to be different, and likewise for eight 8's and one 1. The following program checks the other possibilities: ```DECLARE SUB permute (a\$) DATA 777777722, 666666333, 666666221, 555554444, 555553331, 444433322 CLS FOR tp = 1 TO 6: READ s\$ sod = 0 FOR i = 1 TO 9 sod = sod + VAL(MID\$(s\$, i, 1)) NEXT PRINT s\$; sod sh\$ = s\$ DO s1\$ = LEFT\$(s\$, 3) s2\$ = MID\$(s\$, 4, 3) s3\$ = RIGHT\$(s\$, 3) IF s1\$ <> s2\$ AND s2\$ <> s3\$ AND s1\$ <> s3\$ THEN s4\$ = MID\$(s\$, 1, 1) + MID\$(s\$, 4, 1) + MID\$(s\$, 7, 1) IF s4\$ <> s1\$ AND s4\$ <> s2\$ AND s4\$ <> s3\$ THEN s5\$ = MID\$(s\$, 2, 1) + MID\$(s\$, 5, 1) + MID\$(s\$, 8, 1) IF s5\$ <> s1\$ AND s5\$ <> s2\$ AND s5\$ <> s3\$ AND s5\$ <> s4\$ THEN s6\$ = MID\$(s\$, 3, 1) + MID\$(s\$, 6, 1) + MID\$(s\$, 9, 1) IF s6\$ <> s1\$ AND s6\$ <> s2\$ AND s6\$ <> s3\$ AND s6\$ <> s4\$ AND s6\$ <> s5\$ THEN s7\$ = MID\$(s\$, 1, 1) + MID\$(s\$, 5, 1) + MID\$(s\$, 9, 1) IF s7\$ <> s1\$ AND s7\$ <> s2\$ AND s7\$ <> s3\$ AND s7\$ <> s4\$ AND s7\$ <> s5\$ AND s7\$ <> s6\$ THEN s8\$ = MID\$(s\$, 3, 1) + MID\$(s\$, 5, 1) + MID\$(s\$, 7, 1) IF s8\$ <> s1\$ AND s8\$ <> s2\$ AND s8\$ <> s3\$ AND s8\$ <> s4\$ AND s8\$ <> s5\$ AND s8\$ <> s6\$ AND s8\$ <> s7\$ THEN PRINT sod; END IF END IF END IF END IF END IF END IF permute s\$ LOOP UNTIL s\$ = sh\$ NEXT tp END SUB permute (a\$) DEFINT A-Z x\$ = "" FOR i = LEN(a\$) TO 1 STEP -1 l\$ = x\$ x\$ = MID\$(a\$, i, 1) IF x\$ < l\$ THEN EXIT FOR NEXT IF i = 0 THEN FOR j = 1 TO LEN(a\$) \\ 2 x\$ = MID\$(a\$, j, 1) MID\$(a\$, j, 1) = MID\$(a\$, LEN(a\$) - j + 1, 1) MID\$(a\$, LEN(a\$) - j + 1, 1) = x\$ NEXT ELSE FOR j = LEN(a\$) TO i + 1 STEP -1 IF MID\$(a\$, j, 1) > x\$ THEN EXIT FOR NEXT MID\$(a\$, i, 1) = MID\$(a\$, j, 1) MID\$(a\$, j, 1) = x\$ FOR j = 1 TO (LEN(a\$) - i) \\ 2 x\$ = MID\$(a\$, i + j, 1) MID\$(a\$, i + j, 1) = MID\$(a\$, LEN(a\$) - j + 1, 1) MID\$(a\$, LEN(a\$) - j + 1, 1) = x\$ NEXT END IF END SUB ``` The resulting output is ```777777722 53 666666333 45 666666221 41 555554444 41 555553331 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 444433322 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 ``` Only two sets of digits, 555553331 and 444433322, allow for the formation of eight different 3-digit numbers. The former, however, has a sum of digits equal to 35, which is in fact formable from the digits of itself, and therefore cannot count. Thus the digits used are 444433322, whose sum is the answer: 29. This value is shown 276 times, indicating there are 276 ways of arranging these so that all eight 3-digit numbers as defined are different. The ways are: ```223 224 224 224 224 232 232 233 233 233 233 233 233 233 234 234 234 234 234 234 434 343 344 434 434 434 434 424 434 434 444 444 444 444 244 344 423 424 424 432 443 443 334 334 343 344 443 434 244 424 234 324 423 432 433 324 443 334 343 443 234 234 234 234 234 234 234 234 242 242 242 243 243 243 243 243 243 243 243 243 433 434 434 442 442 443 444 444 343 343 434 234 323 334 342 342 343 343 423 424 442 243 342 334 433 423 233 332 344 443 334 443 444 244 344 434 424 442 443 433 243 243 243 243 243 243 244 244 244 244 244 244 244 244 244 244 244 244 244 244 432 432 433 433 434 434 234 323 324 324 332 334 334 334 334 334 342 342 343 343 434 443 244 442 423 432 343 344 343 433 344 243 324 342 423 432 343 433 342 423 244 244 322 323 323 324 324 324 324 324 324 324 324 324 324 324 324 324 324 324 432 434 434 434 434 244 342 342 343 343 344 344 344 423 432 432 433 433 434 434 343 323 344 244 442 334 434 443 424 442 243 423 432 443 344 443 244 442 234 243 324 324 324 324 324 324 324 324 332 332 332 332 332 332 334 334 334 334 334 334 434 442 442 443 443 443 444 444 424 424 444 444 444 444 242 242 243 243 244 244 342 334 433 234 243 423 233 332 434 443 234 324 423 432 434 443 424 442 234 243 334 334 334 334 334 334 334 334 334 334 342 342 342 342 342 342 342 342 342 342 244 244 244 422 423 424 424 424 442 442 234 243 243 323 324 343 343 344 423 423 324 423 432 443 442 234 243 342 324 432 344 434 443 444 443 244 424 423 344 443 342 342 342 342 342 342 342 342 342 342 342 342 342 343 343 343 343 343 343 343 424 424 432 432 432 433 434 434 434 442 442 443 443 234 234 243 243 423 423 424 334 433 344 434 443 442 324 423 432 334 433 324 423 244 442 244 442 244 442 234 343 343 343 343 343 344 344 344 344 344 344 344 344 344 344 344 344 344 344 344 424 432 432 434 434 224 234 234 234 234 234 242 243 323 324 324 324 324 342 343 432 244 442 224 422 433 243 324 342 423 432 433 423 244 243 342 423 432 243 242 344 344 344 344 344 344 422 422 422 423 423 423 423 423 423 423 423 423 423 423 343 423 424 432 434 434 334 343 434 244 323 334 342 343 344 432 432 433 433 434 422 243 233 342 232 322 344 344 343 433 444 244 344 424 243 344 443 244 442 243 423 423 423 423 423 423 424 424 424 424 424 424 424 424 432 432 432 432 432 432 442 442 443 443 444 444 234 342 343 343 343 343 343 434 234 243 323 343 344 344 334 433 234 432 233 332 334 433 234 243 324 342 423 332 344 434 444 424 324 423 432 432 432 432 432 432 432 432 432 432 433 433 433 433 433 433 433 433 433 433 423 423 424 424 433 434 442 442 444 444 234 242 243 244 244 244 244 244 423 424 344 434 343 433 244 423 334 433 233 332 244 434 244 234 243 324 423 432 244 243 433 433 433 433 433 433 433 434 434 434 434 434 434 434 434 434 434 434 442 442 424 432 432 434 434 442 443 234 234 242 242 243 243 324 342 342 424 424 234 234 432 244 424 242 422 423 422 243 342 334 433 342 423 234 234 243 233 332 334 343 442 442 442 442 442 442 442 442 442 442 442 442 442 442 442 442 443 443 443 443 323 324 324 334 334 334 334 342 342 343 432 432 433 433 434 434 233 234 234 234 443 334 343 234 324 342 423 334 343 243 334 343 324 342 323 332 442 243 324 423 443 443 443 443 443 443 443 443 443 443 443 443 444 444 444 444 243 243 323 324 324 343 343 432 432 433 434 434 323 323 323 323 324 342 442 234 243 224 242 234 243 224 223 232 234 243 324 342 ``` From Enigma No. 1584, "Trisquare", by Adrian Somerfield, New Scientist, 27 February 2010, page 26.

 Subject Author Date re(3): A possible solution Daniel 2010-05-08 01:05:47 re(2): A possible solution Ady TZIDON 2010-05-07 17:39:12 No problem ed bottemiller 2010-05-07 14:07:37 re: A possible solution Daniel 2010-05-07 14:05:34 A possible solution Ady TZIDON 2010-05-07 13:21:12

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