In

this thread, some of Pay The Rent's analysis was examined, but I'm moving over here since I think it has more in common with this thread.

With

this post, I showed that it was impossible to only have one correct solution with the least expensive item in the mailbox. The question I then posed was: What is the minimum number of possible solutions (out of 180) you would need to allow in order to have at least one solution with the least expensive item in the mailbox? Based on the previous post, this must be greater than or equal to two.

Let's examine a Pay The Rent setup with the following prices for the prizes: $21, $16, $11, $6, $4, and $3. Although these numbers were chosen for a particular reason, it will hopefully become clear as the analysis is discussed. We are going to start with the attic, and work our way down to the mailbox.

Clearly, the $21 prize has to go in the attic, as all possible winning solutions have the most expensive prize in the attic. For the second floor, note that when the $16 prize is combined with any other, the total is more than the combination of any other two prizes. For this reason, the $16 prize

**must** belong on the second floor so that, when combined with one other prize, it is more expensive than the first floor. Since the $21 prize is in the attic, the prize that it must be combined with must be less than $5.

There are two possible choices: the $4 prize and the $3 prize. This is important because putting the $4 prize in this slot will allow the $3 to drop down to the mailbox. If the $16 prize was increased to, say, $17.50, it would force the $3 prize to the second floor and not be free for the mailbox. Therefore, two different choices for this slot is the least amount possible.

Placing the $4 prize on the second floor leaves us with $11, $6, and $3 for the first floor and mailbox. So $21//$16 + $4//$11 + $6//$3 is a winning combination, as well as $21//$16 + $4//$11 + $3//$6. Luckily, $21//$16 + $4//$6 + $3//$11 is not a solution, and so we have the minimum possible number of solutions here based on my previous work.

However, we also need to consider that the $3 prize could be placed on the second floor, leaving us with $11, $6, and $4 for the first floor and mailbox. So $21//$16 + $3//$11 + $6//$4 is a winning combination, as well as $21//$16 + $3//$11 + $4//$6. Luckily, $21//$16 + $3//$6 + $4//$11 is not a solution, and so we have the minimum possible number of solutions here as well based on my previous work.

In conclusion, placing the $3 prize on the second floor yields two possible solutions, and placing the $4 prize on the second floor yields two different possible solutions (including the one with the cheapest item in the mailbox). Therefore, we would need to allow

**four** solutions (out of 180) in order to have at least one solution with the least expensive item in the mailbox.

Feel free to tear apart the above as you see fit.