Analytical Question #6

You have twelve coins which look alike and are of equal weight except for one, which may be heavier or lighter. How can you find the odd coin and if it is heavier or lighter by using a balance and only three weighings?



W = Weighing
Create 3 groups with 3 coins each: G1(ABCD), G2(EFGH), G3(IJKL)


I. W1 ==> G1[ABCD] <> G1 (lighter) or G2 (heavier) has the odd coin
Move G2[EFG] to G4, G1[ABC] to G2, and G3[IJK] to G1
Groups: G1[IJKD], G2[ABCH], G3[L], G4[EFG]
I.A. W2 ==> G1[IJKD] <> G1[D] or G2[H] is the odd coin
I.A.1 W3 ==> G1[D] <> G1[D] is the odd coin and is lighter
I.A.2 W3 ==> G1[D] = G3[L] ==> G2[H] is the odd coin and is heavier
I.A.3 W3 ==> G1[D] > G3[L] ==> Not Possible
I.B. W2 ==> G1[IJKD] = G2[ABCH] ==> G4[EFG] has the odd coin and is heavier
I.B.1 W3 ==> G4[E] <> G4[F] is the odd coin
I.B.2 W3 ==> G4[E] = G4[F] ==> G4[G] is the odd coin
I.B.3 W3 ==> G4[E] > G4[F] ==> G4[E] is the odd coin
I.C. W2 ==> G1[IJKD] > G2[ABCH] ==> G2[ABC] has the odd coin and is lighter
I.C.1 W3 ==> G2[A] <> G2[A] is the odd coin
I.C.2 W3 ==> G2[A] = G2[B] ==> G2[C] is the odd coin
I.C.3 W3 ==> G2[A] > G2[B] ==> G2[B] is the odd coin
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II. W1 ==> G1[ABCD] = G2[EFGH] ==> G3[IJKL] has the odd coin
II.A. W2 ==> G1[ABC] <> G3[IJK] has the odd coin and is heavier
II.A.1 W3 ==> G3[I] <> G3[J] is the odd coin
II.A.2 W3 ==> G3[I] = G3[J] ==> G3[K] is the odd coin
II.A.3 W3 ==> G3[I] > G3[J] ==> G3[I] is the odd coin
II.A. W2 ==> G1[ABC] = G3[IJK] ==> G3[L] is the odd coin
II.A.1 W3 ==> G3[I] <> G3[L] is heavier
II.A.2 W3 ==> G3[I] = G3[L] ==> Not Possible
II.A.3 W3 ==> G3[I] > G3[L] ==> G3[L] is lighter
II.A. W2 ==> G1[ABC] > G3[IJK] ==> G3[IJK] has the odd coin and is lighter
II.A.1 W3 ==> G3[I] <> G3[I] is the odd coin
II.A.2 W3 ==> G3[I] = G3[J] ==> G3[K] is the odd coin
II.A.3 W3 ==> G3[I] > G3[J] ==> G3[J] is the odd coin
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III. W1 ==> G1[ABCD] > G2[EFGH] ==> G1 (heavier) or G2 (lighter) has the odd coin
Move G2[EFG] to G4, G1[ABC] to G2, and G3[IJK] to G1
Groups: G1[IJKD], G2[ABCH], G3[L], G4[EFG]
III.A. W2 ==> G1[IJKD] <> G2[ABC] has the odd coin and is heavier
III.A.1 W3 ==> G2[A] <> G2[B] is the odd coin
III.A.2 W3 ==> G2[A] = G2[B] ==> G2[C] is the odd coin
III.A.3 W3 ==> G2[A] > G2[B] ==> G2[A] is the odd coin
III.B. W2 ==> G1[IJKD] = G2[ABCH] ==> G4[EFG] has the odd coin and is lighter
III.B.1 W3 ==> G4[E] <> G4[E] is the odd coin
III.B.2 W3 ==> G4[E] = G4[F] ==> G4[G] is the odd coin
III.B.3 W3 ==> G4[E] > G4[F] ==> G4[F] is the odd coin
III.C. W2 ==> G1[IJKD] > G2[ABCH] ==> G1[D] or G2[H] is the odd coin
III.C.1 W3 ==> G1[D] <> Not Possible
III.C.2 W3 ==> G1[D] = G3[L] ==> G2[H] is the odd coin and lighter
III.C.3 W3 ==> G1[D] > G3[L] ==> G1[D] is the odd coin and heavier