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.

It was so tempting to make a chart for this, but I decided to try and tough it out. It took me a while but the lightbulb finally turned on =)

Since each child has at least one of every marble, and no child has more than 5 of any marble, and no two children have the same number of any type of marble, this means that there must be 1, 2, 3, 4, and 5 of each kind of marble amongst the kids. Now, it doesn’t say that a child can’t have the same number of two types of marble.

Ok, so the first clue gives us a bit of information.

   Name  Age  ag  al  i  m  s
a: J 
b:       10
c:                2 
d:                    3 
e:                       4

Clue 9 says the kid with 3 immies (kid D) can only have 1 or 2 steelies. But Clue 10 says the kid with 5 immies has 1 steelie. Therefore the kid D must have 2 steelies.

Then from Clue 8 we know that Kirby can’t be A (because that’s Josephine) C or D (because they have 2 of some marbles). Therefore Kirby can only be B or E.

Clue 6 says Iris can’t have an even number of aggies, alleys or steelies. Therefore Iris can’t be A (because that’s Josephine) C or D (because they have 2 alleys or 2 steelies). Therefore Iris can only be B or E.

So Clue 11 says that at least 2 kids have twice as many alleys as immies. As it turns out, there are only two kids that could have an even number (2 or 4) of alleys. So there is a kid with 2 alleys and 1 immie, and there is another kid who has 4 alleys and 2 immies. The kid with 2 alleys and 1 immie is kid C.

The kid with 4 alleys and 2 immies can’t be Iris because she can’t have an even number of alleys. S/he can’t be Kirby either because he can’t have 2 of any marbles. So s/he can’t be kid B or E. S/he also can’t be kid C since kid C has 2 alleys and 1 immie. And s/he can’t be kid D because kid D has 3 immies.

Therefore Josephine has 4 alleys and 2 immies. And here’s what we have so far:

   Name  Age  ag  al  i  m  s
a: J              4   2
b: KI    10
c: LH             2   1 
d: HL                 3     2
e: IK                    4

From Clue 10 we know one kid has 3 aggies, 5 immies and 1 steelie. That kid can’t be A, C, or D due to the number of immies. So s/he can only be B or E who are Kirby and Iris. Well, s/he can’t be Iris because Iris has to have an even number of immies. Therefore Kirby has 3 aggies, 5 immies and 1 steelie.

If Kirby were kid E, who also has 4 mibs, that would force the number of alleys to be 2 (since Clue 2 said everyone’s total was 15). Kirby can’t have 2 of any marbles, so Kirby must be kid B. This leaves Iris as kid E. Also, by elimination, kid E (Iris) must have 4 immies.

Now notice Clue 3 and 7. These ones messed me up at first. For some reason I assumed one boy had 5 aggies and the other had 5 steelies. I never considered that one could have both 5 aggies and 5 steelies. That lead me to many confusing contradictions for a while. Anyways, we see that Kirby has neither 5 aggies nor 5 steelies. Therefore Henry must both the boy with 5 aggies and the boy with 5 steelies. Kid D has 2 steelies, so that means Henry must be kid C. And again, since Clue 2 said everyone’s total is 15, this means Henry must have 2 mibs. Also, this means Louise must be kid D.

Going back to Clue 9 we see that Louise can only have 1 or 2 mibs. But Henry already has 2 mibs, so Louise must have 1 mib. Therefore she must have a total of 9 aggies and alleys. The only way to have 9 of only two kinds of marbles is to have 5 of one and 4 of the other. She can’t have 5 aggies and 4 alleys due to Henry and Josephine. Therefore she must have 4 aggies and 5 alleys.

Just in case you are trying to keep up, here is a snap shot of what we know so far.

Name  Age  ag  al  i  m  s
J              4   2
K     10   3       5     1
H          5   2   1  2  5
L          4   5   3  1  2
I                  4  4

The only number of aggies left are 1 and 2. Iris can’t have an even number of aggies, so she must have 1 and Josephine must have 2.

Clue 5 says the 11 year old has 1 aggie, so Iris must be 11.

Clue 4 says there is a kid who has a number of steelies that is exactly half his/her age. That kid must be an even number of years. Between 7 and 11, the only even years are 10 and 8. So we either have a 10 year old with 5 steelies (not the case because Kirby is 10 with 1 steelie) or an 8 year old with 4 steelies. Only Josephine and Iris don’t have their steelies finalized yet. Well, it can’t be Iris who has 4 steelies and is 8 because we already know she is 11. Therefore Josephine is 8 and has the 4 steelies.

This leave Iris with 3 steelies. This means we know she must have had 3 alleys in order to have 15 marbles. By elimination we know Kirby had 1 alley, and in order to have 15 marbles he must have 5 mibs. And by elimination that means Josephine had 3 mibs.

Clue 3 said the boy with 5 steelies was older than at least one other kid. So Henry is not 7. Therefore he must be 9, which makes Louise 7. And here’s the final answer =)

Name  Age  ag  al  i  m  s
J     8    2   4   2  3  4
K     10   3   1   5  5  1
H     9    5   2   1  2  5
L     7    4   5   3  1  2
I     11   1   3   4  4  3

Posted by nikki on 2005-04-22 22:00:58