Not so crowded families

Logic Level 4

In a restaurant, there are a number of families each made from a wife, a husband and their children.

Try to determine how many families and how many boys and girls are in the restaurant knowing the following facts.

a) The number of the children is greater than those of adults.

b) There are more adults than boys.

c) The number of boys is greater than the number of girls.

d) The number of the girls is greater than the number of families.

e) Each family has at least one child.

f) There is no family with the same number of children.

g) Each girl has at least one brother.

h) Each girl has at most one sister.

i) One family has more children than the sum of the number of children of the other families considered anyway.

Enter your answer as $F \times B^G +1$ , where F , B , G are the number of families , boys and also of girls anyway.

Of course it is considered that the husband and wife are adults.

The answer is 1876.

1 solution

Noel Lo
Jul 12, 2016

If we have F families, of course there will be 2F adults (husbands and wives). With more adults than boys, we have 2F > B. With more boys than girls, we have B > G. With more girls than families, we have G > F. Combining we have 2F > B > G > F. By observation, $2F-F$ must be at least 3. In other words, $F$ is at least 3. Suppose $F=3$ . Then $B=5$ while $G=4$ . This means we have 9 children in all spread among 3 families. This is possible with all the conditions fulfilled - we can divide up the 9 children such that we have 6+2+1 or 5+3+1.

Hence $3*5^4+1=1875+1=\boxed{1876}$

@A A . It is really a nice problem. Took a while to get it but loved it! Seeing such questions always gives me inspiration!

Noel Lo - 4 years, 11 months ago

Haha , thanks. It's not my original problem though but thought it's nice as well so considered some others would enjoy it so I posted it.

The problem can be viewed formally I suppose by the relation of restraints , by the way these restraints correlate and as such to find a synthetic proof is pretty hard. By synthetic proof I mean one which takes in only one expression all of the conditions presented which sometime are a little bit heteregeneous in some oints.

By this making a general proof , one which gets all those things in a so to say synthetic expression is pretty hard as well as you observed in your proof where you verified once , the conditions respected by one thing which gives you the synthetic inequality and then the restrictions related to the distributions of kids. Observe further that by this argument you find a solution yet , do not know if there aren't any other solutions to the problem and what I love about it is exactly this synthetic unity over the restraints which is just felt or seen at first in very unclear way.

Can it be possible to obtain a proof where you prove that there is only one solution by a more general treatment of the restrictions ? That's a very intersting thing to ponder I suppose anyway from time to time so think at it as it is like trying to obtain a synthetic expression of the constraints (that is anyway a synthetic unity) which is maybe the source of fascination in your thinking like it is in mine anyway.

A A - 4 years, 11 months ago

Actually, 6+2+1 for the 9 children is impossible. That way we can never get 4 girls.

Saya Suka - 4 years, 7 months ago

maximum no of girls per family can be 2 only because a sister can have at most 1 sister

yasseen N - 1 year, 9 months ago

Not so crowded families

The answer is 1876.

1 solution

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