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.

(In reply to Another program by Charlie)

Thanks, Charlie.

I had made the judgment that my VB program would solve the myriad name/gender/marble combinations, and prove that the one combination was unique, a lot faster than I could do that manually, whereas I could solve the age dimension faster manually (it took me less than 60 seconds) than I could create and execute a subroutine(s) in the program to do it.

But as they say, all roads lead to Rome.

Thanks for taking the time to respond to my post. 

Posted by Penny on 2005-04-24 05:16:19