Two women and two men entered a cafeteria and each pulled a ticket (There's a illustration, but I can't post it here... It's a rectangle with numbers from 5-100 that go by 5 cents, 10 cents, 15 cents...etc)
Anyway...

1. The four ordered the same food and had he same numeral punched out on their ticket (the retangle is the ticket).

2. Each of the four had exactly four coins.

3. The two women had the same amount of money in coins, though no denomination (value) of coin was held by both women; the two men had the same amount of money in coins, though no denominationof coin was held by both men.

4. Each of the four paid the exact amount inndicated by the numberal that was punched out on his or her ticket.

Which numeral was punched out on each ticket? NOTE: "Coins" may be pennies (1 cent), nickles (5 cents), dimes (10 cents), quarters (25 cents), half dollar (50 cents) or silver dollar.

1. All four paid the same price.
2. That price was between 5 cents and 100 cents, and was divisible by 5.

3. Each had 4 coins
4. Each had the same total as one of the others.
5. In each case, the pair with the same amount had no coins in common.
6. So no one had pennies, since his or her "partner" would also need to have pennies.

First assumption (may need to be discarded later) Each person had two kinds of coins.

A: nickels and dimes give 25, 30, 35
B: nickels and quarters give 40, 60, 80
C: nickels and half-dollars give 65, 110, 155
D: dimes and quarters give 55, 70, 85
E: dimes and half-dollars give 80, 120, 160
F: quarters and half-dollars give 125, 150, 175

Shared value 80 (1 nickel & 3 quarters/3 dimes & 1 half-dollar)

We need 2 shared values, so we have to discard the assumption:

G: nickels only gives 20
H: dimes only gives 40
I: quarters only gives 100
J: half-dollars only gives 200

K: nickels, dime, quarters gives 45, 50, 65
L: nickels, dimes, half-dollars gives 70, 75, 115
M: dimes, quarters, half-dollars gives 95, 110, 135

Shared values are: 40 (4 dimes/ 3 nickels & 1 quarter), 65 (3 nickels & 1 half-dollar/1 nickel, 1 dime, & 2 quarters), 70 (2 dimes & 2 quarters/ 2 nickels, 1 dime, & 1 half-dollar), and 110 (2 nickels & 2 half-dollars/1 dime, 2 quarters, & 1 half-dollar)

The 65, 70, and 110 are eliminated because it would be necessary for both "partners" to have at least one coin in the same denomination.

Therefore 40 is the only possibility left.

The problem states that each was able to pay the price in exact change. That would be possible if the price were 30 cents (3 dimes/1 nickel & 1 quarter). Assuming that the men each began with 80 cents, and the women with 40 cents (based on the prices, the problem is old enough to make that assumption about the puzzlemaker's perceptions), One man left with 1 half dollar, the other with 2 quarters; one woman left with 1 dime, the other with 2 nickels.



Personal Note: I remember that the lunch counter at the 5&10 had this "lottery" type of pricing. The tickets with the prices were in balloons. You'd pick a balloon and the waitress would "pop" it. There were probably one or two "free lunch" tickets, several for various amounts between 5 cents and a dollar, and many (half or more) "full price" tickets.

Posted by TomM on 2002-12-27 15:39:59