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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 evelopes
numbered 1 through 15. Each envelope contains the least
number of bills possible of any available
US currency (i.e. no two tens in place of a twenty).
At the auction he makes a successful bid of $8322 for a car.
He hands the auctioneer envelopes
number(s) 2, 8, and 14. After opening the envelopes the
auctioneer finds exactly the right amount.
How many ones did the auctioneer find in the envelopes?
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Answer Posted / guest
Each envelope contains the money equal to the 2 raised to
the envelope number minus 1. The sentence
"Each envelope contains the least number of bills possible
of any available US currency" is only to
misguide you. This is always possible for any amount !!!
One more thing to notice here is that the man must have
placed money in envelopes in such a way
that if he bids for any amount less than $25000, he should
be able to pick them in terms of
envelopes.
First envelope contains, 20 = $1
Second envelope contains, 21 = $2
Third envelope contains, 22 = $4
Fourth envelope contains, 23 = $8 and so on...
Hence the amount in envelopes are $1, $2, $4, $8, $16, $32,
$64, $128, $256, $512, $1024, $2048,
$4096, $8192, $8617
Last envelope (No. 15) contains only $8617 as total amount
is only $25000.
Now as he bids for $8322 and gives envelope number 2, 8 and
14 which contains $2, $128 and $8192
respectively.
Envelope No 2 conrains one $2 bill
Envelope No 8 conrains one $100 bill, one $20 bill, one $5
bill, one $2 bill and one $1 bill
Envelope No 14 conrains eighty-one $100 bill, one $50 bill,
four $10 bill and one $2 bill
Hence the auctioneer will find one $1 bill in the envelopes.
| Is This Answer Correct ? | 6 Yes | 1 No |
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