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## Wednesday, July 30, 2008

### Cash for a Car

Thanks to Lucas Jones we can present you the following puzzle:

A man is going to an Antique Car auction. All purchases must be paid for in cash. He goes to the bank and draws out \$25,000.

Since the man does not want to be seen carrying that much money, he places it in 15 envelopes numbered 1 through 15. Each envelope contains the least number of bills possible of any available US currency (for example, no two tens instead of a twenty).

At the auction he makes a successful bid of \$8322 for a car. He hands the auctioneer envelopes 2, 8, and 14. After opening the envelopes the auctioneer finds exactly the right amount.

The Question: How many ones did the auctioneer find in the envelopes?

Unfortunately, Lucas Jones didn't provide a more elaborate explanation. But thanks to Ronald A. Laski we can now present a more acceptable explanation:

Envelope #2 has 1 \$2 bill in it, which is its total amount also.

Now \$8192 + \$128 + \$2 = \$8322. And that is the winning bid!
The amount of money in each envelope is 2 ^ (# of envelope - 1), except for envelope 15, which contains \$8617.

#### 1 comment:

Anonymous said...

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