Question 1206603: 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 80 stamps, with how many stamps does Ben start?
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
Let the numbers of stamps Alice, Ben, and Carl had originally be A, B, and C, respectively.
A B C originally
A-B-C 2B 2C after Alice gave each of Ben and Carl the number they had
2A-2B-2C -A+3B-C 4C after Ben gave each of Alice and Carl the number they had
4A-4B-4C -2A+6B-2C -A-B+7C after Carl gave each of Alice and Ben the number they had
In the end, they each had 80 stamps:
(1) 4A-4B-4C=80
(2) -2A+6B-2C=80
(3) -A-B+7C=80
Simplifying (1) and (2)...
(4) A-B-C=20
(5) -A+3B-C=40
Adding (3) and (4) to eliminate A:
(6) -2B+6C=100
Adding (4) and (5) to eliminate A:
(7) 2B-2C=60
Adding (6) and (7) to eliminate B:
(8) 4C=160; C=40
Plugging C=40 in (7) to solve for B:
2B-80=60; 2B=140; B=70
Plugging C=40 and B=70 in (4) to solve for A:
A-70-40=20; A=130
ANSWER: Ben started with B=70 stamps.
CHECK:
initially: (A,B,C) = (130,70,40)
after first exchange: (20,140,80)
after second exchange: (40,40,160)
after third exchange: (80,80,80)
Answer by ikleyn(52800)  (Show Source):
You can put this solution on YOUR website! .
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 80 stamps, with how many stamps does Ben start?
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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 = 80,
2b = 80,
c-a-b = 80.
Their solution is a = 80/2 = 40, b = 80/2 = 40, c = 80 + a + b = 80 + 40 + 40 = 160.
Thus, immediately before step 3 and after step 2
Alice has 40 stamps, Ben has 40 stamps, Carl has 160 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 = 40,
b-a-c = 40,
2c = 160.
Their solution is a = 40/2 = 20, c = 160/2 = 80, b = 40 + a + c = 40 + 20 + 80 = 140.
Thus, immediately before step 2 and after step 1
Alice has 20 stamps, Ben has 140 stamps, Carl has 80 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 = 20,
2b = 140,
2c = 80.
Their solution is b = 140/2 = 70, c = 80/2 = 40, a = 20 + b + c = 20 + 70 + 40 = 130.
Thus, immediately before step 1 (initially)
Alice has 130 stamps, Ben has 70 stamps, Carl has 40 stamps.
ANSWER. Ben starts with 70 stamps.
CHECK. Step 1: (130, 70, 40) ---> (130-70-40, 140, 80) = (20, 140, 80).
Step 2: (20, 140, 80) ---> ( 40, 140-20-80, 160) = (40, 40, 160).
Step 3: (40, 40, 160) ---> ( 80, 80, 160-40-40) = (80, 80, 80). ! correct !
Solved by the backward method without using complicated equations.
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