The 25 Weirdest Interview Questions

Diego Calderon

Grad Student
2k - 1

Say you have 4 condoms; first guy puts all of them on, second guy uses the top one again, third guy uses the bottom one inverted, and then you have the two that were in between that can be used 2 times each. That's 1+1+1+2*2 = 7 men or 2*4-1, etc etc

Of course, this doesn't protect the woman from the (possible) STDs of the men! Just the men are protected...sorry ladies lol

I rushed my first answer, today's been too effin busy!

peterruse
 
its actually k + k/2 :D

That's just k.

It's 2k-1 where k>0, one side is for the woman, otherwise every side can be used by a man using the general technique

A. Take 1 condom W to be used on the woman, then nest the remaining k-1, one at a time, separating afterwards and inserting a new one. (k-1)
B. One guy takes her down with W. (1)
C. k-1 rounds again, this time inverted and nested in W (k-1)

k-1+1+k-1= 2k-1 | k>0
 
Yep, it's 2k-1 :)

Now I guess you can generalize to (w) women, (k) condoms...

Edit: Actually, a better formulation would be: Given (m) men and (w) women, what's the minimum number of condoms needed if each man fancies every woman?
 
7. Out of 25 horses, pick the fastest 3 horses. In each race, only 5 horses can run at the same time. What is the minimum number of races required? (Reportedly from Bloomberg LP)
Run 5 original races with all of the horsies
Run the winners of all 5 races, pick the winner as the fastest horse
Run the 4 other original winners, as well as the 2nd place of the group with the fastest horse, pick winner as 2nd fastest
Run 4 others still in that top group and: 3rd place of the 1st place horse's original group if the 2nd place horse won, or the 2nd place of the 2nd place horse's original group if one of the 4 other groups won. Pick winner as 3rd fastest.

Your solution seems to add to eight races---I do believe you can do it in seven.

Assuming, of course, that you have no stopwatch to measure times and so you can only qualitatively see who is 1st, 2nd, 3rd, etc. in a five-horse race. Also assume no ties.

(i) Organize into groups of five and run five races, like you said. For example, in Group X, Label the winners and the groups they are from, in order, as X, Xa, Xb, Xc, Xd

(ii) For the sixth race, run winners of the five groups, like you said. For example, in Group X, Relabel the winners and the groups they are from, in order, as 1, 2, 3, 4, 5.

(iii) Now, and because we only care about the fastest three horses and because we only have five groups, we can write out all the possibilities for the fastest three horses--They are:

1 1a 1b
1 1a 2
1 2 1a
1 2 2a
1 2 3

You will see that there are only five possible candidates for the 2nd and 3rd fastest horse--- 1a, 1b, 2, 2a, and 3.

(iv) Race these five together in the seventh race. We know that the winner of this race will be 1a or 2, and we must see who the 1st and 2nd places winners of this race is to determine the second and third fastest horses (just looking at who the winner of the 7th race is not enough).
 
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