Five neighborhood children (two of whom Henry and Kirby are boys, and three of whomIris, Josephine, and Louiseare girls) are counting their marbles. Each child is a different age (7 through 11 years) and has at least one but no more than 5 of each of aggies, alleys, immies, mibs, and steelies. No two children have the same number of the same type of marble. From the information provided determine each child's age and the number of each type of marble in his or her collection.
1. The five children are Josephine, the 10 year old, the child who owns 2 alleys, the girl who has 3 immies, and the child who has 4 mibs.
2. Each child owns exactly 15 marbles.
3. The boy who has five steelies is older than at least one other child.
4. The number of steelies in one child's collection is exactly half his or her age.
5. The 11 year old child has one aggie.
6. The only types of marble of which Iris has an even number are immies and mibs.
7. One of the boys has 5 aggies.
8. Kirby doesn't own 2 of any type of marble.
9. The child with 3 immies has fewer than 3 mibs and fewer than 3 steelies.
10. One child has 3 aggies, 5 immies, and 1 steelie.
11. Each of at least 2 children has exactly twice as many alleys as immies.
(In reply to
In response to overwhelming demand...... by Penny)
This one does take age into consideration:
DECLARE SUB permute (a$)
age$ = "789ab" ' order hkijl
ag$ = "12345"
al$ = "12345"
im$ = "12345"
mi$ = "12345"
st$ = "12345"
OPEN "allmarbl.txt" FOR OUTPUT AS #2
FOR agen = 1 TO 120
PRINT agen;
IF MID$(age$, 4, 1) <> "a" THEN
FOR aln = 1 TO 120
ix = INSTR(al$, "2")
IF ix <> 4 AND ix <> 2 AND MID$(age$, ix, 1) <> "a" THEN
FOR imn = 1 TO 120
ix = INSTR(im$, "5")
im5 = ix
im1 = INSTR(im$, "1")
im2 = INSTR(im$, "2")
IF MID$(al$, im1, 1) = "2" AND MID$(al$, im2, 1) = "4" THEN good = 1: ELSE good = 0
ix = INSTR(im$, "3")
im3 = ix
IF (ix = 3 OR ix = 5) AND im2 <> 2 AND good = 1 THEN
IF MID$(age$, ix, 1) <> "a" AND MID$(al$, ix, 1) <> "2" THEN
FOR min = 1 TO 120
ix = INSTR(mi$, "4")
IF MID$(age$, ix, 1) <> "a" AND MID$(al$, ix, 1) <> "2" AND MID$(im$, ix, 1) <> "3" AND MID$(mi$, 2, 1) <> "2" AND MID$(mi$, im3, 1) < "3" THEN
FOR agn = 1 TO 120
ix = INSTR(ag$, "5")
IF ix < 3 THEN good = 1: ELSE good = 0
IF MID$(ag$, 2, 1) = "2" THEN good = 0
IF MID$(ag$, im5, 1) <> "3" THEN good = 0
ix = INSTR(ag$, "1")
IF MID$(age$, ix, 1) = "b" AND good = 1 THEN
FOR stn = 1 TO 120
good = 0
FOR child = 1 TO 5
IF MID$(age$, child, 1) = "8" AND MID$(st$, child, 1) = "4" THEN good = 1
IF MID$(age$, child, 1) = "a" AND MID$(st$, child, 1) = "5" THEN good = 1
NEXT child
IF MID$(st$, 2, 1) = "2" THEN good = 0
IF MID$(st$, im3, 1) > "3" THEN good = 0
IF MID$(st$, im5, 1) > "1" THEN good = 0
IF INSTR("24", MID$(im$, 3, 1)) = 0 THEN good = 0
IF INSTR("24", MID$(mi$, 3, 1)) = 0 THEN good = 0
IF INSTR("24", MID$(ag$, 3, 1)) > 0 THEN good = 0
IF INSTR("24", MID$(al$, 3, 1)) > 0 THEN good = 0
IF INSTR("24", MID$(st$, 3, 1)) > 0 THEN good = 0
ix = INSTR(st$, "5")
IF ix < 3 AND MID$(age$, ix, 1) > "7" AND good = 1 THEN
FOR child = 1 TO 5
tot = VAL(MID$(ag$, child, 1)) + VAL(MID$(al$, child, 1)) + VAL(MID$(im$, child, 1)) + VAL(MID$(mi$, child, 1)) + VAL(MID$(st$, child, 1))
IF tot <> 15 THEN good = 0: EXIT FOR
NEXT child
IF good THEN
PRINT : PRINT age$: PRINT ag$: PRINT al$: PRINT im$: PRINT mi$: PRINT st$
PRINT #2, : PRINT #2, age$: PRINT #2, ag$: PRINT #2, al$: PRINT #2, im$: PRINT #2, mi$: PRINT #2, st$
END IF
END IF
permute st$
NEXT stn
END IF
permute ag$
NEXT agn
END IF
permute mi$
NEXT min
END IF
END IF
permute im$
NEXT imn
END IF
permute al$
NEXT aln
END IF
permute age$
NEXT agen
CLOSE
SUB permute (a$)
DEFINT AZ
x$ = ""
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
giving the answer as:
9ab87
53124
21345
15423
25431
51342
where the sequence is Henry, Kirby, Iris, Josephine and Louise.
The first row is for age, where a represents 10 and b represents 11. Subsequent rows show number of aggies, alleys, immies, migs and steelies.

Posted by Charlie
on 20050423 23:28:10 