Math in the modern world

Question

Alan, Ben and Carl collect NBA cards. They exchange cards among

themselves. According to the following scheme: Alan gives Ben as many

cards as Ben has and Carl as many cards as Carl has. After that, Ben gives

Alan and Carl as many cards as each of them has, and then Carl gives Alan

and Ben as many cards as each has. If each finally has 64 cards, with how

many cards does each have at the beginning?(work backwards)

Answer 1

Working backwards....

At the end, (A,B,C) = (64,64,64)

notice there are always 192 cards total... we use the fact that prior to each transfer of cards, the 'other' two people had 1/2 has many cards, and the sum of those two quantities plus the prior 'donor's card count must be 192:

Prior step, A,B had half as many, C had 192 - (A+B): (A,B,C) = (32,32,128)

Prior step, A,C had half as many, B had 192- (A+C): (A,B,C) = (16, 112 ,64)

Prior step, B,C had half as many, A had 192 - (B+C): (A,B,C) = (104 ,56, 32)

Thus, at the start A=104, B=56, C=32

Check:

Start: (104,56,32)

After Alan gives cards: (16, 112, 64)

After Ben gives cards: (32, 32 , 128)

After Carl gives cards: (64, 64, 64)

The problem can be solved working forwards as follows...

Initial numbers: A,B,C (Alan has 'A', Ben has 'B', Carl has 'C')

After Alan givesaway cards: A-B-C, 2B, 2C

After Ben gives away cards: 2(A-B-C), 2B-(A-B-C)-2C, 4C

2(A-B-C), -A+3B-C, 4C

After Carl gives away cards: 4(A-B-C), 2(-A+3B-C), 4C-2(A-B-C)-(-A+3B-C)

4(A-B-C), 2(-A+3B-C), -A-B+7C

Now we can write three equations in three unknowns:

4(A-B-C) = 64

2(-A+3B-C) = 64

-A-B+7C = 64

Simplifying the first two:

A-B-C = 16

-A+3B-C = 32

-A-B+7C = 64

This system has the solution A=104, B=56, C=32 and those check out.