Brainteaser 1742: Not-so-safe deposit
From The Sunday Times, 4th February 1996 [link]
The safe deposit boxes at our bank are arranged in a huge pyramid, the top of which is illustrated:
When the were first opened the top box had a quantity of gold coins placed in it and the rest were empty. One night a burglar broke into that box, emptied it, and (cleverly, he thought) instead of running off with the coins, he found the two empty boxes in the row below and put some of his haul in one and the rest in the other.
Word spread through the underworld about the bank’s lax security, and a sequence of burglaries took place. Each burglar broke into one box containing some coins, emptied it, found two empty boxes in the row below, placed some of his haul in one of these boxes and the rest in the other.
Ages later the manager of the bank opened up the top box and was horrified to find it empty. So he tried the boxes in the next row and found them empty too. He carried on like this until he found some coins in one particular row. In fact, no matter how many burglaries had taken place, there could not have been more empty rows at the top of the pyramid. And indeed if any fewer burglaries had taken place all those rows could not have been empty.
(a) How many empty rows at the top were there?
(b) How many burglaries were there?
This puzzle is included in the book Brainteasers (2002). The puzzle text above is taken from the book.
[teaser1742]
John Crabtree and 
Lise Andreasen and 
GeoffR and 
Frits, 
S2T2 
Jim Randell 1:45 pm on 10 November 2019 Permalink |
We can try to empty the uppermost rows as quickly as possible.
Initially, in the top row there is 1 box, initially containing the complete amount:
The first burglar breaks in, and splits the amount between the two boxes in the second row. So each of the 2 boxes contains 1/2 of the complete amount:
The second burglar breaks in and splits the amount from one of the boxes in the second row between 2 (of the 3) boxes in the third row:
The third burglar can’t burgle the remaining box in the second row, as there aren’t enough empty boxes in the third row to hide the spoils. So he splits the amount from one of the boxes in the third row into 2 (of the 4) boxes in the fourth row:
The fourth burglar can burgle the box in the second row, as there are now 2 empty boxes in the third row:
And so the process continues:
So we’ve managed to empty the first 3 rows (in 14 burglaries).
Can we empty the fourth row?
What if every row below it was full? We would have:
We can check this by calculating the sum the first few terms:
So there is not enough room to hide the entire amount using just rows 5 onwards, hence we can never empty row 4.
Solution: (a) The first 3 rows were empty.
The number of burglaries required to achieve any situation is one less than the number of occupied boxes.
And as the fractions of the original collection decrease as we move down the rows we want to occupy the minimal number of boxes possible. If the first 3 rows are empty this minimal number is 4 on the 4th row, 5 on the 5th row, 6 on the 6th row (4/8 + 5/16 + 6/32 = 1), which required 4 + 5 + 6 − 1 = 14 burglaries.
Solution: (b) There were 14 burglaries.
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John Crabtree 9:50 pm on 13 November 2019 Permalink |
While experimenting with this, I found that one box in row 2 could be replaced by:
i) a pyramid with one box full in row 4, two in row 5, three in row 6 etc
or ii) one box full in row 4, and three in all lower rows
Add the two together, and the grid would be full with two boxes full in row 4, and all all boxes in lower rows. It does not matter if this pattern can actually be reached. Row 4 cannot be empty. it is quite easy to check if the first three rows can be emptied.
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