Bottle cap

[wolfgang]

If you have 6 Bottles weighing(100gm. each,and only one is 99.9gm).

each bottle has a cap, each cap is 5gm and only one of these caps is 5.1gm.

you have no idea which is which.

What are the minimal times of weighing,using a scale balance,in order to find the heavy cap?

you are not allowed to weigh the caps alone,i.e. they should be placed with bottles on the scales.

you can change the caps between the bottles as you want.

[CaptainEd]

Call the bottles B1-B6 and the caps C1-C6

W1) B123, C123 vs. B456, C456

<: Left side has Light Bottle (LB) and RIght side has Heavy Cap (HC)

--W11) B4C4 vs B5C5

--<: C5 is HC

-->: C4 is HC

--=: C6 is HC

>: reverse of (<) case: Left side has HC, one more weighing

=: one side has LB and HC

--W12) B123 C456 vs B456 C123 (swap caps to other side)

--<: C123 has HC, B456 normal bottles

----W121) B4C1 vs B5 C2

----<: C2 is HC

---->: C1 is HC

----=: C3 is HC

-->: reverse of (<) case

--=: won't happen

So the longest chain of weighings is

W1, W12, W121

Edited January 11, 2012 by CaptainEd

[wolfgang]

Call the bottles B1-B6 and the caps C1-C6

W1) B123, C123 vs. B456, C456

<: Left side has Light Bottle (LB) and RIght side has Heavy Cap (HC)

--W11) B4C4 vs B5C5

--<: C5 is HC

-->: C4 is HC

--=: C6 is HC

>: reverse of (<) case: Left side has HC, one more weighing

=: one side has LB and HC

--W12) B123 C456 vs B456 C123 (swap caps to other side)

--<: C123 has HC, B456 normal bottles

----W121) B4C1 vs B5 C2

----<: C2 is HC

---->: C1 is HC

----=: C3 is HC

-->: reverse of (<) case

--=: won't happen

So the longest chain of weighings is

W1, W12, W121

thats right...thanks

[bhramarraj]

If there are two such bottles A, B with caps a, b. Then we may find the required bottle and cap as under, with the conditions given.

Weigh-1: Aa V/S Bb; Weigh-2: Ab V/S Ba;

Results: [1] If Aa = Bb and Ab > Ba, Then B is light and a is heavy.

[2] If Aa = Bb and Ab < Ba, Then A is light and a is heavy.

[3] If Aa > Bb Then a is heavy and B is light. (No need of weigh-2)

If there are three bottles A, B, C with caps a, b, c, then-

Weigh-1: Aa V/S Bb; Weigh-2: Ab V/S Ba;

Results: [1] If Aa = Bb, and Ab = Ba; then C is light and c is heavy.

[2] If Aa = Bb, and Ab > Ba; then B is light and b is heavy.

[3] If Aa > Bb, and Ab = Ba; then B is light and a is heavy.

[4] If Aa > Bb, and Ab > Ba; then B is light and c is heavy.

[5] If Aa > Bb, and Ab < Ba; then C is light and a is heavy.

If there are four Bottles A, B, C, D with caps a, b, c, d, then-

Weigh-1: Aa V/S Bb; Weigh-2: Cc V/S Dd;

Results: [1] If Aa = Bb & Cc = Dd, then Weigh-3: Ac V/S BD

(Note: Ac can’t be equal to BD); so if Ac > BD, then c

is heavy & B is light; and if Ac < BD, then d is heavy &

A is light.

[2] If Aa = Bb & Cc > Dd; Then c is heavy and D is light,

(Weigh-3 is not necessary.)

[3] If Aa = Bb & Cc < Dd; Then d is heavy and C is light

(Weigh-3 is not necessary.)

[4] If Aa > Bb & Cc = Dd; Then a is heavy and B is light

(Weigh-3 is not necessary.)

[5] If Aa < Bb & Cc = Dd; Then b is heavy and A is light

(Weigh-3 is not necessary.)

[6] If Aa > Bb & Cc > Dd; Then Weigh-3: Ac V/S Ca,

if Ac > Ca then c is heavy & B is light,

if Ac < Ca, then a is heavy & D is light.

(Note: Ac can’t be eqal to Ca.)

[7] If Aa > Bb & Cc < Dd; Then Weigh-3: Ad V/S Da,

if Ad > Da then d is heavy & B is light,

if Ad < Da, then a is heavy & C is light.

(Note: Ad can’t be eqal to Da.)

[8] If Aa < Bb & Cc > Dd; Then Weigh-3: Bc V/S Cb,

if Bc > Cb then c is heavy & A is light,

if Bc < Cb, then b is heavy & D is light.

(Note: Bc can’t be eqal to Cb.)

[9] If Aa < Bb & Cc < Dd; Then Weigh-3: BD V/S Db,

if BD > Db then d is heavy & A is light,

if BD < Db, then b is heavy & C is light.

(Note: BD can’t be eqal to Db.)

[bhramarraj]

Let the bottles be A, B, C, D, E & F, all weighing 100gm, except one which weighs 99.9gm, is say L.

Let the caps be a, b, c, d, e & f, all weighing 5gm, except one which weighs 5.1 gm, is say H.

Case-I:

Weigh1: Aa + Bb = Cc + Dd, then both H & L are on the same side. Weigh2: Ae + Db V/S Fa + BD

Results: If Ae + Db = Fa + BD, then e can’t be H, because then either A or D would have to be L for balanced RHS and LHS , but according to weigh-1, A could be L only when either a or b was H. Also D could be L only when either c or d was H. Similarly d can’t be H, and F can’t be L.

So there are three possibilities: (i) b is H, A is L. OR (ii) a is H, B is L. OR (iii) f is H, E is L

Weigh3: Af V/S Eb

Results: If Af = Eb then possibility (ii) above, is right i.e. a is H and B is L.

If Af > Eb then f is H and E is L.

If Af < Eb then b is H and A is L.

Case-II:

Weigh-1: Aa + Bb = Cc + Dd; then both H & L are on the same side.

Weigh-2: Ae + Db > Fa + BD;

Results: There are three possibilities (i) b is H, B is L OR (ii) e is H, F is L OR (iii) f is H, E is L

Weigh-3: Bf V/S Eb;

Results: If Bf = Eb, then possibility (ii) above, is right, i. e. e is H andF is L.

If Bf > Eb, then f is H and E is L.

If Bf < Eb, then b is H and B is L.

Similarly we may find the required bottle and cap in three weighing, when Weigh-2: Ae + Db < Fa + BD.

Similarly if Weigh-1: Aa + Bb > Cc + Dd, and if Aa + Bb < Cc + Dd. we may find required bottle and cap in three weighing.