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| All the Marbles (Posted on 2005-04-22) |
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Five neighborhood children (two of whom- Henry and Kirby- are boys, and three of whom-Iris, Josephine, and Louise-are 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.
Solution (VB program / manual logic)
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 | Comment 7 of 18 | 
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I wrote a VB program to perform the tedious logic of combining and eliminating names, genders, and marble counts. (It's a long and laborious program but, thanks to the magic of copy/pasting, it only took a few minutes to write, and it ran in just a few seconds. The bigger my programs are, the faster they run. I won't reproduce it here). This resulted in only one valid combination of those parameters:
Henry: 5 aggies, 2 alleys, 1 immies, 2 mibs, 5 steelies.
Kirby: 3 aggies, 1 alley, 5 immies, 5 mibs, 1 steelies.
Iris: 1 aggie, 3 alleys, 4 immies, 4 mibs, 3 steelies.
Josephine: 2 aggies, 4 alleys, 2 immies, 3 mibs, 4 steelies.
Louise: 4 aggies, 5 alleys, 3 immies, 1 mib, 2 steelies.
Then I performed the age logic manually.
Iris is 11 years old, because she has 1 aggie. One of the children is twice as old as the number of steelies, and since Henry has 2 alleys, he cannot be 10, since hint #1 says that the 10 year old and the owner of 2 alleys are separate people. Therefore Josephine is 8 (2 * 4 steelies). Henry cannot be 7, since he has 5 steelies and so he must be older than somebody; we already know he can't be 10; so Henry is 9. Louise cannot be 10 because she owns 3 immies, and hint #1 says that the 10 year old and the girl who owns 3 immies are two separate people; so Louise is 7. Kirby must be 10.
So my final result is:
Henry (9 yrs): 5 aggies, 2 alleys, 1 immie, 2 mibs, 5 steelies.
Kirby (10 yrs): 3 aggies, 1 alley, 5 immies, 5 mibs, 1 steelies.
Iris (11 yrs): 1 aggie, 3 alleys, 4 immies, 4 mibs, 3 steelies.
Josephine (8 yrs): 2 aggies, 4 alleys, 2 immies, 3 mibs, 4 steelies.
Louise (7 yrs): 4 aggies, 5 alleys, 3 immies, 1 mib, 2 steelies.
Edited on April 23, 2005, 1:51 pm
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Posted by Penny
on 2005-04-23 13:43:47 |
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