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Alice, Ben, and Carl collect stamps. They exchange stamps among themselves according to the following scheme:
(1) Alice gives Ben as many stamps as Ben has and Carl as many stamps as Carl has.
(2) After that, Ben gives Alice and Carl as many stamps as each of them has,
(3) 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?
use polya's four method.
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I will solve the problem backward.
FIRST BACKWARD STEP, inverse to the base step 3
Let "a", "b" and "c" be the numbers of stamps that Alice, Ben and Carl had after step 2 immediately before step 3.
Step 3 is this transformation (a, b, c) ---> (2a, 2b, c-a-b).
So, the description of step 3 gives us these equations
2a = 64,
2b = 64,
c-a-b = 64.
Their solution is a = 64/2 = 32, b = 64/2 = 32, c = 64 + a + b = 64 + 32 + 32 = 128.
Thus, immediately before step 3 and after step 2
Alice has 32 stamps, Ben has 32 stamps, Carl has 128 stamps.
SECOND BACKWARD STEP, inverse to the base step 2
Let "a", "b" and "c" be the numbers of stamps that Alice, Ben and Carl had after step 1 immediately before step 2.
Step 2 is this transformation (a, b, c) ---> (2a, b-a-c, 2c).
So, the description of step 2 gives us these equations
2a = 32,
b-a-c = 32,
2c = 128.
Their solution is a = 32/2 = 16, c = 128/2 = 64, b = 32 + a + c = 32 + 16 + 64 = 112.
Thus, immediately before step 2 and after step 1
Alice has 16 stamps, Ben has 112 stamps, Carl has 64 stamps.
THIRD BACKWARD STEP, inverse to the base step 1
Let "a", "b" and "c" be the numbers of stamps that Alice, Ben and Carl had before step 1 (i.e. initially).
Step 1 is this transformation (a, b, c) ---> (a-b-c, 2b, 2c).
So, the description of step 1 gives us these equations
a-b-c = 16,
2b = 112,
2c = 64.
Their solution is b = 112/2 = 56, c = 64/2 = 32, a = 16 + b + c = 16 + 56 + 32 = 104.
Thus, immediately before step 1 (initially)
Alice has 104 stamps, Ben has 56 stamps, Carl has 32 stamps.
ANSWER. Alice starts with 104 stamps.
CHECK. Step 1: (104, 56, 32) ---> (104-56-32, 112, 64) = (16, 112, 64).
Step 2: (16, 112, 64) ---> ( 32, 112-16-64, 128) = (32, 32, 128).
Step 3: (32, 32, 128) ---> ( 64, 64, 128-32-32) = (64, 64, 64).
Solved by the backward method without using complicated equations.