## 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!

Back to the puzzle
Copyright © 1996-2018. RJE-productions. All rights reserved. No part of this website may be published, in any form or by any means, without the prior permission of the authors.
This website uses cookies. By further use of this website, or by clicking on 'Continue', you give permission for the use of cookies. If you want more information, look at our cookie policy.