[Guest]

If I have a balance scale and 8 spheres, one sphere heavier than the rest, how many times to I have to weigh the spheres in order to insure that I can find the heavier one?

Well, I can put more than two spheres on the scale at a time so, it takes me 3 times to weigh the spheres. 4 to 4, 2 to 2, 1 to 1.

[Guest]

Adding to my puzzle above, here goes another puzzle. If I have 100 red socks in a drawer, 100 black socks in a drawer, 100 blue socks in a drawer, how many socks do I have to take to insure I have a pair? Please answer with the minimum answer.

4 socks. Worst possible scenario I take 1 of each then I take one that matches with the ones I took already.

[plainglazed]

only two weighings are needed to assure finding the heavy sphere.

[Guest]

only two weighings are needed to assure finding the heavy sphere.

1. Balance 3 x 3

2. If they are the same, balance the remaining two and you will have the heavy one

3. If they are not the same, balance two of the heavier three against each other.

4. If they are not the same you will know which is the heavy one.

5. If they are the same, the remaining one is obviously the heavy one.

[Guest]

Assuming you mean that the other 7 all weigh the same, it would take 3 weighings to find the heavy one, weighing 4 vs 4 to find which 4 has the heavy one, then weigh 2 vs 2 and 1 vs 1. I can not think of a more clever way to weigh this one that uses less than 3. But if the other 7 do not weigh the same, that complicates matters and the above method will not work.In that case I think it would take 7 weighings, weighing 1 sphere vs 1 sphere at a time and always keeping the heavy one for the next weigh.

Reading the above spoilers I see someone did find a more clever way to weigh!

4 socks will ensure at least 1 pair, either getting 1 pair and two of the other color or 2 pair.

Edited May 1, 2011 by Nana7

[Guest]

2

Let me complicate this question.

Now there are 2 heavier balls and 6 normal ones. How many weighings would it take to identify both heavier balls in the worst case scenario?

If you are done with 2 then try with 3 heavy balls, then 4.(total 8 balls only)

Need not try with 5/6/7 heavy balls coz the answer would be same as 3/2/1 heavy balls respectively.

Edited May 1, 2011 by Mogli

[plainglazed]

haven't given this much thought...yet. suspect there may be a differnce if the two "heavies" weigh the same or not. might I suggest starting another topic for this thread. almost missed it as i had already seen the original. looks interesting.