Solution to: Cash for a Car

The solution: one.

Unfortunately, Lucas Jones did not provide a more elaborate explanation. However, thanks to Ronald A. Laski we can now present a more acceptable explanation:

In the envelopes numbered 1 up to 15, the man placed the following amounts of money: $1, $2, $4, $8, $16, $32, $64, $128, $256, $512, $1024, $2048, $4096, $8192, $8617. The amount of money in each envelope is 2^(number of envelope - 1), except for envelope 15, which contains $8617.

In envelope number 14, which contains $8192, there are:

    81 $100 bills = $8100
     1 $ 50 bill  = $  50
     2 $ 20 bills = $  40
     1 $  2 bill  = $   2
                    ----- +
                    $8192

In envelope number 8, which contains $128, there are:

     1 $100 bill  = $100
     1 $ 20 bill  = $ 20
     1 $  5 bill  = $  5
     1 $  2 bill  = $  2
     1 $  1 bill  = $  1 <- that is the one!
                    ----
                    $128

Envelope number 2, which contains $2, has one $2 bill in it.

Now $8192 + $128 + $2 = $8322, which is the winning bid!

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