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 Count them (Posted on 2010-11-22)
In how many ways will two WRONGs make one RIGHT?

 No Solution Yet Submitted by Ady TZIDON No Rating

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 computer solution Comment 2 of 2 |

DEFDBL A-Z
FOR w = 1 TO 9
used(w) = 1
FOR r = 1 TO 9
IF used(r) = 0 THEN
used(r) = 1
FOR o = 0 TO 9
IF used(o) = 0 THEN
used(o) = 1
FOR n = 0 TO 9
IF used(n) = 0 THEN
used(n) = 1
FOR g = 0 TO 9
IF used(g) = 0 THEN
used(g) = 1

wrong = w * 10000 + r * 1000 + o * 100 + n * 10 + g
test\$ = LTRIM\$(STR\$(2 * wrong))
IF LEN(test\$) = 5 THEN
IF VAL(MID\$(test\$, 1, 1)) = r AND VAL(MID\$(test\$, 3, 1)) = g THEN
i = VAL(MID\$(test\$, 2, 1))
h = VAL(MID\$(test\$, 4, 1))
t = VAL(MID\$(test\$, 5, 1))
IF used(i) = 0 AND used(h) = 0 AND used(t) = 0 THEN
IF i <> h AND i <> t AND h <> t THEN
PRINT wrong, test\$
ct = ct + 1
END IF
END IF
END IF
END IF

used(g) = 0
END IF
NEXT
used(n) = 0
END IF
NEXT
used(o) = 0
END IF
NEXT
used(r) = 0
END IF
NEXT
used(w) = 0
NEXT

PRINT ct

finds what I assume is the expected solution:

` wrong        right 12734        25468 12867        25734 12938        25876 24153        48306 24765        49530 25173        50346 25193        50386 25418        50836 25438        50876 25469        50938 25734        51468 25867        51734 25938        51876 37081        74162 37091        74182 37806        75612 37846        75692 37908        75816 49153        98306 49265        98530 49306        98612 21 solutions in all for the decimal version.`

But, taking this further:

In base 9:

DEFDBL A-Z
FOR w = 1 TO 8
used(w) = 1
FOR r = 1 TO 8
IF used(r) = 0 THEN
used(r) = 1
FOR o = 0 TO 8
IF used(o) = 0 THEN
used(o) = 1
FOR n = 0 TO 8
IF used(n) = 0 THEN
used(n) = 1
FOR g = 0 TO 8
IF used(g) = 0 THEN
used(g) = 1

wrong = w * 6561 + r * 729 + o * 81 + n * 9 + g
test\$ = ""
tst = 2 * wrong
DO
d\$ = LTRIM\$(STR\$(tst MOD 9))
tst = tst \ 9
test\$ = d\$ + test\$
LOOP UNTIL tst = 0

IF LEN(test\$) = 5 THEN
IF VAL(MID\$(test\$, 1, 1)) = r AND VAL(MID\$(test\$, 3, 1)) = g THEN
i = VAL(MID\$(test\$, 2, 1))
h = VAL(MID\$(test\$, 4, 1))
t = VAL(MID\$(test\$, 5, 1))
IF used(i) = 0 AND used(h) = 0 AND used(t) = 0 THEN
IF i <> h AND i <> t AND h <> t THEN
PRINT w; r; o; n; g, test\$
ct = ct + 1
END IF
END IF
END IF
END IF

used(g) = 0
END IF
NEXT
used(n) = 0
END IF
NEXT
used(o) = 0
END IF
NEXT
used(r) = 0
END IF
NEXT
used(w) = 0
NEXT

PRINT ct

` w  r  o  n  g              right 1  2  7  4  6              25603 2  4  1  7  3              48356 3  7  5  4  1              76182just 3 solutions.`

And in base 11:

DEFDBL A-Z

FOR w = 1 TO 10
used(w) = 1
FOR r = 1 TO 10
IF used(r) = 0 THEN
used(r) = 1
FOR o = 0 TO 10
IF used(o) = 0 THEN
used(o) = 1
FOR n = 0 TO 10
IF used(n) = 0 THEN
used(n) = 1
FOR g = 0 TO 10
IF used(g) = 0 THEN
used(g) = 1

wrong = w * 14641 + r * 1331 + o * 121 + n * 11 + g
test\$ = ""
tst = 2 * wrong
DO
d\$ = MID\$("0123456789abcdef", (tst MOD 11) + 1, 1)
tst = tst \ 11
test\$ = d\$ + test\$
LOOP UNTIL tst = 0

IF LEN(test\$) = 5 THEN
r2 = INSTR("0123456789abcdef", MID\$(test\$, 1, 1)) - 1
g2 = INSTR("0123456789abcdef", MID\$(test\$, 3, 1)) - 1
i = INSTR("0123456789abcdef", MID\$(test\$, 2, 1)) - 1
h = INSTR("0123456789abcdef", MID\$(test\$, 4, 1)) - 1
t = INSTR("0123456789abcdef", MID\$(test\$, 5, 1)) - 1
IF r2 = r AND g2 = g THEN
IF used(i) = 0 AND used(h) = 0 AND used(t) = 0 THEN
IF i <> h AND i <> t AND h <> t THEN
PRINT w; r; o; n; g, test\$
ct = ct + 1
END IF
END IF
END IF
END IF

used(g) = 0
END IF
NEXT
used(n) = 0
END IF
NEXT
used(o) = 0
END IF
NEXT
used(r) = 0
END IF
NEXT
used(w) = 0
NEXT

PRINT ct

` w  r  o  n  g              right 1  2  5  3  10             24a79 1  2  7  4  3              25386 1  2  7  10  4             25498 2  4  1  9  3              48376 2  4  1  10  3             48396 2  4  5  0  10             48a19 2  4  5  1  10             48a39 2  4  5  3  10             48a79 2  4  7  5  3              493a6 2  4  8  3  5              4956a 2  4  8  5  6              49601 2  5  7  4  3              50386 2  5  7  6  4              50418 2  5  7  10  4             50498 2  5  8  7  6              50641 2  5  10  1  9             50937 3  7  6  5  1              741a2 3  7  8  1  5              7452a 3  7  8  5  6              74601 3  7  9  6  8              74825 4  9  0  8  1              97152 4  9  1  8  3              97356 4  9  2  6  5              9751a 4  9  3  2  6              97651 4  9  6  5  1              981a2 4  9  7  1  3              98326 4  9  7  5  3              983a6fully 27 solutions. `

Note the successor to 9 is shown as 10 for wrong, but as a for right.

`But 49 are found for base 12: 1  2  3  5  6              246b0 1  2  5  3  10             24a78 1  2  5  7  11             24b3a 1  2  5  9  11             24b7a 1  2  7  8  3              25346 1  2  7  10  3             25386 1  2  7  11  3             253a6 1  2  9  3  6              25670 1  2  10  3  8             25874 1  2  10  7  9             25936 1  2  11  3  10            25a78 1  2  11  4  10            25a98 2  4  1  11  3             483a6 2  4  3  5  6              486b0 2  4  5  6  11             48b1a 2  4  5  7  11             48b3a 2  4  5  9  11             48b7a 2  4  7  10  3             49386 2  4  7  11  3             493a6 2  4  8  6  5              4950a 2  4  11  0  10            49a18 2  4  11  1  10            49a38 2  4  11  3  10            49a78 3  7  4  8  9              72956 3  7  5  0  10             72a18 3  7  5  4  10             72a98 3  7  5  6  11             72b1a 4  9  0  10  1             96182 4  9  0  11  1             961a2 4  9  3  8  7              96752 4  9  5  0  10             96a18 4  9  5  1  10             96a38 4  9  5  3  10             96a78 4  9  5  7  11             96b3a 4  9  6  10  1             97182 4  9  6  11  1             971a2 4  9  8  6  5              9750a 4  9  11  0  10            97a18 4  9  11  1  10            97a38 4  9  11  2  10            97a58 5  11  0  8  1             ba142 5  11  0  9  1             ba162 5  11  1  3  2             ba264 5  11  1  7  3             ba326 5  11  1  8  3             ba346 5  11  2  3  4             ba468 5  11  3  4  6             ba690 5  11  3  6  7             ba712 5  11  4  7  9             ba936 `

It seems the trend is an increasing number the higher the base.

 Posted by Charlie on 2010-11-22 18:15:41

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