Friday Puzzler -- Men's Club

A pie full of blackbirds ...

Larry Barrett

plus six.

  Sorry, but no

Alex Y

but I have never been able to envision the size of a pie that contained 24 blackbirds!

Hint

Alex Y

Work with the negatives of each condition.

Hint #2

Alexy

If the number of people meeting all four conditions is as small as possible, what does that say about the number of people who fail at least one condition?

Re: Hint #2

Larry Barrett

Ten.

From the problem statement,

1. 85 have hair, 15 do not have hair

2. 70 have land lines phones, 30 do not have land line phones

3. 80 are married, 20 are not married

4. 75 have homes, 25 do not have homes

Pick any set of three, say 1, 2, and 4. There are a total of 15+30+25=70 that are in the "not have" set. So of the remaining group 3, 70 of the 80 married men could line up with these 70 "not haves". The remaining 10 married will have to overlap with all the "haves".

Not sure of this reasoning, but it works for any combination of 3 of 4 sets.

Re: Hint #2

Alex Y

Correct. The "at least one failure" set is maximized if the individual failure sets are disjoint. So members

1-15 are bald; the rest have hair

16-45 use only their cell phones; the rest have land lines

46-65 are single; the others are married, and

66-90 live in multi-family residences; the others in single family.

So only members number 91-100 meet all four of the initial conditions.