A calculus for a puzzle

Edward Tsang 2015.12.04

Here is a puzzle for fun. A systematic approach ensures that results are easily verifiable.

The Puzzle:

  • $2 buys one bottle of soft drink.
  • 4 caps can be used to exchange for one new bottle.
  • 2 bottles can be used to exchange for one new bottle.
    • How many bottles of soft drink can one acquire with $10?

    A calculus for solving the above puzzle

    Let A be the number of soft drinks acquired. Let B and C be bottles and caps respectively.
    This is the calculus:
    $2 --> (1A, 1B, 1C)
    4C --> (1A, 1B, 1C)
    2B --> (1A, 1B, 1C)
    (aA, bB, cC) + (a'A, b'B, c'C) = (a+a'A, b+b'B, c+c'C)

    Application to the above puzzle

    $10 --> (5A, 5B, 5C) --> (5+1A, 5+1B, 5-4+1C) = (6A, 6B, 2C)
    --> (6+3A, 6-6+3B, 2+3C) = (9A, 3B, 5C)
    --> (9+1A, 3+1B, 5-4+1C) = (10A, 4B, 2C)
    --> (10+2A, 4-4+2B, 2+2C) = (12A, 2B, 4C)
    --> (12+1A, 2+1B, 4-4+1C) = (13A, 3B, 1C)
    --> (13+1A, 3-2+1B, 1+1C) = (14A, 2B, 2C)
    --> (14+1A, 2-2+1B, 2+1C) = (15A, 1B, 3C)

    Answer: $10 gets one 15 bottles of soft drink.

    Source of the puzzle

    I was told that this question was used in a "TSA" exam in Hong Kong -- an exam which primary 3 students (8 year old?) take.

    I suspect children may solve the puzzle in a different way. I wouldn't be surprised if they find the answer quicker than I do. I try not to make mistakes, in case the Hong Kong government strips me of my primary school qualification...

    [End]