(In reply to
re: computer findings--unambiguously by Ady TZIDON)
All those sets with 9 as the highest value are shown:
5 6 6 9 9
4 6 6 9 9
3 6 6 9 9
2 6 6 9 9
1 6 6 9 9
5 5 6 9 9
4 5 6 9 9
3 5 6 9 9
2 5 6 9 9
1 5 6 9 9
4 4 6 9 9
3 4 6 9 9
2 4 6 9 9
1 4 6 9 9
3 3 6 9 9
2 3 6 9 9
1 3 6 9 9
2 2 6 9 9
1 2 6 9 9
1 1 6 9 9
4 5 5 9 9
3 5 5 9 9
2 5 5 9 9
1 5 5 9 9
4 4 5 9 9
3 4 5 9 9
2 4 5 9 9
1 4 5 9 9
3 3 5 9 9
2 3 5 9 9
1 3 5 9 9
2 2 5 9 9
1 2 5 9 9
1 1 5 9 9
3 4 4 9 9
2 4 4 9 9
1 4 4 9 9
3 3 4 9 9
2 3 4 9 9
1 3 4 9 9
2 2 4 9 9
1 2 4 9 9
1 1 4 9 9
2 3 3 9 9
1 3 3 9 9
2 2 3 9 9
1 2 3 9 9
1 1 3 9 9
1 2 2 9 9
1 1 2 9 9
5 5 6 6 9
4 5 6 6 9
3 5 6 6 9
2 5 6 6 9
1 5 6 6 9
4 4 6 6 9
3 4 6 6 9
2 4 6 6 9
1 4 6 6 9
3 3 6 6 9
2 3 6 6 9
1 3 6 6 9
2 2 6 6 9
1 2 6 6 9
1 1 6 6 9
4 5 5 6 9
3 5 5 6 9
2 5 5 6 9
1 5 5 6 9
4 4 5 6 9
3 4 5 6 9
2 4 5 6 9
1 4 5 6 9
3 3 5 6 9
2 3 5 6 9
1 3 5 6 9
2 2 5 6 9
1 2 5 6 9
1 1 5 6 9
3 4 4 6 9
2 4 4 6 9
1 4 4 6 9
3 3 4 6 9
2 3 4 6 9
1 3 4 6 9
2 2 4 6 9
1 2 4 6 9
1 1 4 6 9
2 3 3 6 9
1 3 3 6 9
2 2 3 6 9
1 2 3 6 9
1 1 3 6 9
1 2 2 6 9
1 1 2 6 9
4 4 5 5 9
3 4 5 5 9
2 4 5 5 9
1 4 5 5 9
3 3 5 5 9
2 3 5 5 9
1 3 5 5 9
2 2 5 5 9
1 2 5 5 9
1 1 5 5 9
3 4 4 5 9
2 4 4 5 9
1 4 4 5 9
3 3 4 5 9
2 3 4 5 9
1 3 4 5 9
2 2 4 5 9
1 2 4 5 9
1 1 4 5 9
2 3 3 5 9
1 3 3 5 9
2 2 3 5 9
1 2 3 5 9
1 1 3 5 9
1 2 2 5 9
1 1 2 5 9
3 3 4 4 9
2 3 4 4 9
1 3 4 4 9
2 2 4 4 9
1 2 4 4 9
1 1 4 4 9
2 3 3 4 9
1 3 3 4 9
2 2 3 4 9
1 2 3 4 9
1 1 3 4 9
1 2 2 4 9
1 1 2 4 9
2 2 3 3 9
1 2 3 3 9
1 1 3 3 9
1 2 2 3 9
1 1 2 3 9
1 1 2 2 9
There are 140 of these.
Added to the 75 previously found, where the largest value was 6, this makes 215 possible 5-digit numbers.
The new ones were found with:
CLS
d5 = 9
nct(d5) = nct(d5) + 1
FOR d4 = d5 TO 1 STEP -1
IF nct(d4) < 2 AND (d4 = 9 OR d4 <= 6) THEN
nct(d4) = nct(d4) + 1
FOR d3 = d4 TO 1 STEP -1
IF nct(d3) < 2 AND (d3 = 9 OR d3 <= 6) THEN
nct(d3) = nct(d3) + 1
FOR d2 = d3 TO 1 STEP -1
IF nct(d2) < 2 THEN
nct(d2) = nct(d2) + 1
FOR d1 = d2 TO 1 STEP -1
IF nct(d1) < 2 THEN
nct(d1) = nct(d1) + 1
PRINT d1; d2; d3; d4; d5
numCt = numCt + 1
IF numCt MOD 40 = 0 THEN
DO: LOOP UNTIL INKEY$ > ""
PRINT
END IF
nct(d1) = nct(d1) - 1
END IF
NEXT
nct(d2) = nct(d2) - 1
END IF
NEXT
nct(d3) = nct(d3) - 1
END IF
NEXT
nct(d4) = nct(d4) - 1
END IF
NEXT
PRINT numCt
|
Posted by Charlie
on 2010-12-23 03:33:13 |