## 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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