I apologize for leaving you without a puzzle last weekend. I went away, joining my husband at a conference in the South of France and meeting our long-time friends from LA. Friends we haven't seen for 12-18 years, friends that got married, had kids, some divorced.
Image by Cayusa, distributed under CCL.
Bear with me, this leads to a puzzle, but I have to explain first. On this specific weekend, my husband and I found ourselves in the company of 4 men and 12 kids. The wives and ex-wives all went on a long planned weekend trip to Paris. As appropriate in such situations, the men tried to group together for survival. They all met at the house of one of them in a countryside, they brought wine, cheeses, baguette and a lot of chocolate. They managed impressively well, with the older kids taking care of the younger, with everyone above age 10 parting till 2am and everyone having a great time.
There were 4 men, each a father of 3 kids.
One of them left two older kids at home and brought two friends of his daughter instead.
Among the kids there were 2 pairs of twins of different ages, each pair of the same gender.
One men was laughed at by his friends that his wife and he go to Greece and make kid on a different island each year.
Two fathers had only boys.
For each pair of twins there was a kid of the same age but of opposite gender from another family.
You now have to sort kids by families.
There were:
a 3-year -old boy,
a 4-year-old boy ,
two 5-year-old boys,
a 7-year -old girl,
two 11-year-old boys and one 11-yer-old girl,
three 13-year-old girls and one 13-year-old boy.
I do not use actual photographs for privacy reasons; the image above is just for illustration and should not be used as a hint. So, who brought what kids?
Your answers are accepted any time until midnight Eastern Time on Sunday, on our Family Puzzle Marathon.
10 comments:
Here is one answer (there are others which are variations. The variations cannot be separated out because nothing more is said about the younger children.
F1 . . . F2 . . F3. . . F4
11BT . . 11G. . 13TGP . 13B
11BT . . 5B. . 13TGP . .4B
7B . . . 5B. . 13G . . .3B
Notes
=====
F means Father
T means Twin
B means Boy
G means Girl
P means pal (unrelated friend of 13 old girl of F3)
The five year olds cannot be twins: there is no 5 year old girl to balance them. I have put them in the same Family to ensure that only 1 family has different age children (F4) for their children. It seems to me however, that two 5 year non twins would get some sort of comment.
Since the comment did not come, only 1 family has different age children. If you really want avoid the 5 year old problem, interchange the 5 year old boy in family 2 with 7 year olds in family 1. That leaves you with 2 families of different ages.
Two fathers have nothing but boys-- F1 and F4
Each set of twins is the same gender. F1 has 11 year old boys F3 has 3 kids with him -- his daughter and her 2 pals. The two friends are the other set of twins. They would likely be the same age. Teen age girls don't wander too far from that rule.
A clarifying comment - the list of kids above contains ages of all the kids present in this house on this night. This means that there are 3 sets of 3 siblings and, from the 4th family, a daughter and her two friends.
You need to figure out what dad brought what kids.
The first firm constraint is that we must have 3 kids 1 year apart, which means 3yo boy, 4yo boy and 5yo boy form the first group.
Secondly, I assume that a girl with two friends would all be the same age. That can either be the three 11-yo, or some combination of the four 13-yo. But because we have to form another all-boys group, we can't have the two 11-yo boys with a girl in the same group. We also can’t have two 13 yo girls and a 13yo boy since that rules out the second pair of twins. So that means that the second group is the three 13-yo girls, two of them twins.
From here, we actually have two alternative solutions that both satisfy all the conditions, the difference being the 5yo boy and 13yo boy alternating between the remaining groups.
A. Third group (and a second boys-only group) is twin 11yo boys plus a 5yo boy. Fourth group is 7yo girl, 11yo girl and a 13yo boy.
B. Third group (and a second boys-only group) is twin 11yo boys plus a 13yo boy. Fourth group is 7yo girl, 11yo girl and a 5yo boy.
I hate doing this as an alternate answer to the one I gave, but it is possible. The clue I'm having the most problem with is that only 1 father took it on the chin for having 3 children of different ages.
I'm using the same notation as before.
F1: 11BT 11BT 4B
F2: 11G 3B 7G
F3: 13GTF 13GTF 13G (daughter)
F4: 13B 5B 5B
F1 and F4 have boys
F3 brought two girls who are friends of his daughter and twins.
F2 is the only one who had kids of different ages.
F4: is the sticker. The five year old boys cannot be twins for 2 reasons: there is no girl who is 5 in the company and we need to make them born in (say) January and December but they are still 5 I think. Anyway no two fathers can have children of different ages. May I suggest that one of the 5 year olds is adopted? Perhaps their birthdays are 1 day apart.
Family 1: Boys ages 3, 4, and 5. (the father who had a child every year on a different Greek island. Also a family with all boys)
Family 2: 11 yr. old twin boys and a 13 yr. old boy. (Another family of all boys.)
Family 3: 13 yr. old twin girls and a 5 yr. old boy
Family 4: 13 yr. old girl and her 11 yr. old girl friend and 7 yr. old girl friend. (or daughter is 11 with 13 yr. old and 7 yr. old friends.)
Family 1 seemed obvious if I read the clue correctly. And the 11 yr. old and 13 yr. old twins were obviously in different families since there are 3 kids per family. With the makeup of the children remaining, Family 2 had to be made up of all boys. At this point I think you can have two choices: the girl with the friend has all friends of one gender or she has friends close to her age but not of the same gender. If she has friends of the same gender then Family 2 has a 13 yr. old boy, Family 3 has a 5 yr. old boy and Family 4 has a 13 yr.old girl, 11 yr.old girl and 7 yr.old girl. If Family 4 has friends by age them Family 2 has a 5 yr. old boy, Family 3 has a 7 yr.old girl and Family 4 has 13 yr. old girl with a 13 yr. old boy friend and 11 yr. old girl friend.
Family one:
3-yr old boy, 4-yr old boy, 5-yr old boy
(family that had kids each yr after trip to Greece; family with only boys)
Family two:
11-yr old boy, 11-yr old boy (twins)
5-yr old boy
(family with only boys)
Family three:
13-yr old girl, 13-yr old girl (twins)
7-yr old girl
Family four:
13-yr old girl, 11-yr old girl, 13-yr old boy
(girl with two friends)
TracyZ
Ok, the actual families present were:
Father 1: 3, 4 and 5 year old boys. This is the father that made kid every year on a different Greek island.
Father 2: 13 year old daughter and her two friends, 13-year old twin girls. I know this constrain is not very strong but at teen age kids usually group by age and gender when coming over to a place where parents are present. It is most natural for a 13 year old girl to bring two of her girlfriends.
Father 3: 11 year old twin boys and a 13 year old boy. Another dad with only boys.
Father 4: 5 year old boy, 7 year old girl and 11 year old girl.
Ilya correctly caught me up in some ambiquity - third father may have a 5-year old boy instead of a 13 year old boy producing a boy family as well. I guess the last constraint should have been that one twin family has a kid of the same age as another twin family.
You all deserve a puzzle point. Have a great week!
I'm still having problems with this. Father 4 in your solution could get the same comment as Father 1. (The couple go away each year to conceive a child). A restriction is not put on that makes the kids 1 year apart.
Jerome - to your last comment. I think that the following constraint does mean that the couple's kids are one year apart: "One men was laughed at by his friends that his wife and he go to Greece and make kid on a different island each year."
So, only kids aged 3, 4 and 5 could fit.
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