Card Swap

[miamivince]

Chas, Mick & Sam swap cards over 3 rounds. Round 1, M and S have their # of cards doubled by C. In the next 2 rounds, M and then S double the # of cards held by the other 2. If they now have 24 cards each, how many did M start with ? I assume each started with c, m,s. This summed to 72. c>m>s .

After round 1 the totals are c-m-s, 2m, 2s . After round 3 , I got 4c-4m-4s, 2m-2c-6s, -s -3m-3c . The answer is 21. The last tally for round 3 is clearly wrong.
Would it be faster to use a geometric progression formula ? Thank you. No workings given with the soln.

 

[HallsofIvy]

Chas, Mick & Sam swap cards over 3 rounds. Round 1, M and S have their # of cards doubled by C. In the next 2 rounds, M and then S double the # of cards held by the other 2. If they now have 24 cards each, how many did M start with ? I assume each started with c, m,s. This summed to 72. c>m>s .

Was the fact that they summed to 72 given?

After round 1 the totals are c-m-s, 2m, 2s .

Okay.

After round 3 , I got 4c-4m-4s, 2m-2c-6s, -s -3m-3c .

What happened to round two? After round two C and S are doubled and M is reduced by those numbers. The numbers are 2(c- m- s)= 2c-2m-2s, 2m- (c-m-s)- 4s= 3m-c+ s ,4s. Then after round three C and M are doubled and S is reduced: 2(2c-2m- 2s)= 4c- 4m- 4s, 2(3m- c+ s)= 6m- 2c+ 2s, 4s- (2c-2m- 2s)- (3m- c+ s)= 5s- m+3c

The answer is 21. The last tally for round 3 is clearly wrong.

Yes, it is because (4c- 4m- 4s)+ (2m- 2c- 6s)+ (-s- 3m- 3c)= -c- 5m- 7s, not "c+m+s" as it clearly should be since the total number of cards does not change.