how many stamps does alice start

Question

Alice, Ben and Carl collect stamps. They exchange stamps among themselves according to the following scheme: Alice gives Ben as many stamps as Ben has and Carl as many stamps as Carl has. After that, Ben gives Alice and Carl as many stamps as each of them has, and then Carl gives Alice and Ben as many stamps as each has. If each finally has 64 stamps, with how many stamps does Alice start?

Answer 1

At the start Alice, Ben, and Carl have A,B,and C stamps, respectively.

 

On the first transaction, Alice gives B stamps to Ben and C stamps to Carl.

So after the first transaction

  Alice has A-B-C

  Ben has 2B

 Carl has 2C

 

 

On the second transaction, Ben gives  A-B-C stamps to Alice and 2C stamps to Carl.

 

Alice has 2(A-B-C) = 2A-2B-2C

 

Ben has 2B - (A-B-C)-2C = 2B - A +B+C = -A + 3B -C

 

Carl has 4C

 

 

On the final transaction Carl gives  2A-2B-2C stamps to Alice and  -A+3B-C stamps to Ben

 

Alice has 4A-4B-4C stamps.

 

Ben has -2A+6B-2C stamps

 

Carl has  4C - (2A-2B-2C)-(-A+3B-C) =

             4C -2A + 2B + 2C + A - 3B + C =

             -A  - B+7C 

 

Each of these is 64 stamps, so....

 

Alice:

4A-4B-4C =64

 A-B-C = 16 <--- this is equation #1

 

 

Ben:

-2A+6B-2C = 64

  -A + 3B - C = 32<--- divides by 2; this is equation #2

 

Carl: -A - B+7C  = 64 <--- this is equation #3

 

 

The 3x3 system of equations is:

A-B-C = 16

-A + 3B  -C = 32

-A - B+7C = 64

 

Pairing up the first to of these 3 equations, A is eliminated by adding together equations #1 and #2:

  2B - 2C = 48

   B - C = 24

   B = C+24  <---- please label this equation ALPHA

 

Next SUBTRACTS Equation #3 MINUS equation #2, which also eliminates A:

8c - 4b = 32

2c - b = 8

 

2c - (C+24) = 8

2c - c - 24 = 8

c - 24 = 8

 c = 32

 

By equation ALPHA in bold above, B = 32+24 = 56

Finally, using the ORIGINAL first equation A-b-c = 16

                                                             a - 56 - 32 = 16

                                                             a - 88 = 16

                                                             a = 104

 

So altogether , they starts with A+B+C = 104+56+32 = 192 stamps.

At the end, there are also 3 x 64 = 192 stamps.

 

Finally, reviewing the transactions:

 

A=104

B=56

C=32

 

Alice gives 56 stamps to Ben and 32 stamps to Carl:

A=16

B=112

C=64

 

 

Ben gives 16 stamps to Alice and 64 stamps to Carl:

A=32

B=32

C=128

 

Finally Carl gives 32 stamps to both Alice and Ben:

A=64

B=64

C=64

 

Which shows the correct answer of:

A=104
B=56
C=32

 

Answer 2

If A_n, B_n, and C_n are the numbers of stamps Alice, Ben and Carl have after n exchanges for n=0,1,2,3 then initial values are A_o, B_o, and C_o.

After the first exchange

A₁=A_o-B_o-C_o,

B₁=2B_o and,

C₁=2C_o.

After the second exchange,

A₂=2A₁,

B₂=B₁-A₁-C₁, and

C₂=2C_1.

After the third exchange,

A₃=2A_2,

B₃=2B₂, and

C₃=C₂-A₂-B₂ and

A₃=B₃=C₃=64.

64=2A₂=2B₂ → A ₂=B₂=32

64=C₂-32-32→ C₂=128.

Then A₁=16 and C₁=64 and B₁=32+16+64=112

Then B_o=56 and C_o=32 and A_o, Alice's initial number of stamps, is A_o=16+56+32=104