Three thieves -- Abe, Bob, and Cal -- went to a cave one night. Each carried a bag of stolen gold coins.
During the night, Abe woke up, stole half of Bob's
coins and half of Cal's, added them to his own
bag, and went back to sleep. At this point,Abe
had 300 coins.
A little later, Bob woke up and did the same thing
to Abe and Cal; and an hour after that, Cal woke
up and did the same thing to Abe and Bob.
When each thief checked his bag in the morning, he
discovered that he had the same number of coins he
had brought to the cave.
How many did each thief bring?
Answer to the Problem: Abe brought 75 coins.
Bob brought 150 coins.
Cal brought 300 coins.
Let x = number of coins that Abe brought.
Let y = number of coins that Bob brought.
Let z = number of coins that Cal brought.
-
After Abe steals coins from Bob and Cal,
- Abe has x + y/2 + z/2 which equals 300 coins,
- Bob has y/2 coins, and
- Cal has z/2 coins.
-
After Bob steals coins from Abe and Cal,
- Abe has 150 coins,
- Bob has y/2 + z/4 + 150 coins, and
- Cal has z/4 coins.
-
After Cal steals coins from Abe and Bob,
- Abe has 75 coins,
- Bob has y/4 + z/8 + 75 coins, and
- Cal has z/4 + 75 + y/4 + z/8 + 75 coins.
Therefore, Abe must have brought 75 coins, so x =75.
Since x + y/2 + z/2 = 300 and since x = 75,
it follows that y + z = 450.
Since the number of coins that Bob brought equals the
number of coins that he wound up with,
y = y/4 + z/8 + 75
Multiplying by 8,
we get 8y = 2y + z + 600
or 6y - z = 600.
Solving these two equations simultaneously,
y + z = 450 and
6y - z = 600,
we get y = 150 and z = 300.
Correctly solved by:
| 1. Jia Ran | Rome, Italy |
| 2. Tom Marino | Winchester, VA |
| 3. Richard Mocarski | Winchester, VA |
| 4. Chris Moats | Winchester, VA |
| 5. Matthew Reames | Roanoke, VA |
| 6. Chip Crawford | Winchester, VA |
| 7. Court Pifer | Winchester, VA |
| 8. Jon Pence | Winchester, VA |