[Guest]

you have x number normal balls, out of which 2 are heavy. you have a standard balance scale for which to weigh the balls. what is the maximum balls you can have, assuming you want to identify the two heavy balls in only four weightings.

if you can find a way to garentee that you get the two heavy in 4 weightings with 10 total balls, your reasonably intelligent.

12 balls your brilliant, 13 balls your a genius.

[bonanova]

you have x number normal balls, out of which 2 are heavy. you have a standard balance scale for which to weigh the balls. what is the maximum balls you can have, assuming you want to identify the two heavy balls in only four weightings.

3⁴ = 81 outcomes of 4 weighings.

₂C₁₃ = 78, while ₂C₁₄ = 91.

No more than 13 balls.

I'll do the weighing sequence later today if no one has it.

[Guest]

edit... my spacing didnt take for some reason so i tried to spread it out with line break to help

hey im sure people have answered this and then not have posted it because its a lil lengthy but im in transit so ive got nothing better to do

So the balls are named ABCDEFGHIJKLM respectively also to simplify some parts name the first 5 balls G1(group 1) the next 4 G2 and the last 4 G3

the numbers are what compare your on, just to help organize thoughts.

if > means the first one mentioned min compare (or take) is greater then the second im not actually saying if > 1, im saying if >, 1 .....

1 so take G2 and compare to G3

if = you either have one in G2 and one in G3 or two in G1

2 take FGH and compare to EIJ

3 if = take D to J

4 if = 2 in ABC so your... done

4 if > 1 in D one in ABC... done

4 if < 1 in J one in FGH...done

3 if < take BCI to ELF (by now u know F is empty so it just makes it even balls)

4 if > 1 in I and JKM...done

4 if < 1 in E and one in A or D...done

4 if = 1 in I one in L or one int E one in BC ...done

3,4 if > one in FGH one in KLM...done

1 if one is > then it doesnt matter which for the process so just rename whichever one is heavier with G2's letters

so if G3 is bigger it is renamed FGHI

2 take BCF to GH (with a low weight ball from the smaller group)

3 if = compare AB to CD

if = then one in F one in GH or one in I and e

4 compare G to H and as you can see ...done

if > one in A one in I or one in B one in Gh

4 compare GH...done

if < one in C one in GH. or d and I

4 compare GH... done

3 if> take BCI to FGH

4 if = one in F one in IBC...done

4 if > one in I one in BC...done

4 if < one in F one in ADE...done

3 if < take AG to DH

4 if = balls in GD or HA or GH so test GH... done

4 if > one in G one in AEI so test AE...done

4 if < one in H one in DEI so test DE...done

.. i havent done example cases to make sure but i have gone through the process and i dont see any errors

this part may be rambling to many of you but

as bonanova similarly said

it should be noted that this can be done by simpley realizing that if you compare 3 to 3 at the beginning you would end with if they are = 30 options > and < 24 so the equal is impossible as you can only eliminate 2/3s of the amount left each time

with 4 to 4 you get =,<,> all 26 options which is less the 3^3 so that can work

5 to 5 you get = 28 <>25 which also doesnt work because 28>27

so you need to start with 4 to 4 and then use similar logic to find that the rest

anyway good waste of a lil of a car and plane ride

Edited June 28, 2009 by final

[bonanova]

I was distracted by personal business last week and worked on this only through the first weighing.

Agree the first weighing has to be 4 vs 4 -- each case, < = > gives 26 cases.

I could not get beyond that, however.

Particularly having two possible locations for the heavy balls after the first weighing,

and then to limit the cases for each outcome to 9 and 3 for the next two.

I'll look over final's approach, but it seems he's got it.

Great puzzle and solution both!

I also wondered about the easier cases of 10, 11 and 12 balls.

First thought is that for 10 balls, your first weighing is 3 vs 3.