Two-part Analysis (TPA) | Re: Four friends, Jamilah, Kermit, Madison, and Parker, trade three types

Let Aggies be A, Swirlies be S and Tom Bowlers be T
Given:
1S can be exchanged for 2A
1T can be exchanged for 2S

Which also means that 1T can be exchanged for 4A

Current Holdings

Jamilah: 2T, 1A
Kermit: 1A, 1S
Madison: 2S, 1T
Parker: 4A, 1S

1st trade (Between Jamilah and Parker):
Jamilah gives 1T in exchange for 4A

Current Holdings after 1st trade

Jamilah: 1T, 5A
Kermit: 1A, 1S
Madison: 2S, 1T
Parker: 1T, 1S

2nd trade (Between Jamilah and Madison):
Jamilah gives 1T in exchange for 2S

Current Holdings after 2nd trade

Jamilah: 2S, 5A
Kermit: 1A, 1S
Madison: 2T
Parker: 1T, 1S

Now, the third trade has to be any single marble exchange for the same value marble between any two valid marble holders. Since 7 is the highest option, and Jamilah already has 7, we would not consider Jamilah as a part of the third trade.

At this point, the maximum number of marbles is 7 and the minimum is 2

Statistics : Posted by sleEZy • on 06 Jul 2024, 08:09 • Replies 2 • Views 56

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