Non-VIP Forever! came up with a very interesting approach to the question:
- Can we deduce the distribution of 3, 5 and 8 Coins from TLE results if we only have the number of Coin Packs and the number of Coins they granted?
The proposed solution was: Each Coin Pack grants at least 3 Coins, so we can multiply the number of Packs with 3 and subtract this from the total number of Coins. The remaining number has to be a combination of 5 or 8 Coins.
If this method always leads to one single solution, i. e. if there is only one possible combination of 3, 5 or 8 Coins that leads to the total, we will be able to generally add TLE results to the statistics database. In the three provided examples, the conjecture holds.
Vice versa, if there was a sum of Coins that can be build by more than one combinations of 3, 5 or 8 Coins, the method would be too much prone to errors. Especially when there are greater amounts of Coins, the calculating person may overlook a possible other combination.
Unfortunately, while fiddling around with some numbers, I already found two examples with more than one possible solutions.
- 25 Coins from 5 Packs can be a result of $ 5 \cdot 5 $ Coins (column 1a) or $ 3 \cdot 3 + 8 \cdot 2 $ Coins (column 1b).
- Another example where both solutions contain 3-Coin results would be 31 Coins from 7 Packs which can be a result of $ 3 \cdot 2 + 5 \cdot 5 $ Coins (column 2a) or $ 3 \cdot 5 + 8 \cdot 2 $ Coins (column 2b).
So after thinking about it, I will not include data from TLEs as it is quite time-consuming to verify if the calculations are correct.
But I was interested in a more formal solution. Generally, the task is to find if there is more than one solution for
- the same sum of Coins which is represented by $ 3a + 5b + 8c $
- with the same number of Coin Packs wich is which is $ a + b + c $.
More precisely, are there any $ a, b, c \in \N $ and $ x, y, z \in \N $ where at least $ a \ne x $, $ b \ne y $ or $ c \ne z $ that are a solution to the following equation system?
- $ 3a + 5b + 8c = 3x + 5y + 8z $
- $ a + b + c = x + y + z $
As I was lazy, I had a look at the first case of column 1a that had 25 Coins from five 5-Coin results and no 3-Coin or 8-Coin results at all. I generalized this to all results made up of only 5-Coin results $ b $ with no 3-Coin or 8 Coin results, that is, I set $ a = 0 $ and $ c = 0 $ which makes the equation system easier to handle:
- $ 5b = 3x + 5y + 8z $
- $ b = x + y + z $
Solving the first equation for $ b $ delivers
- $ b = \tfrac35x + y + \tfrac85z $
- $ b = x + y + z $.
By equating the right hand sides of the equations, we obtain
- $ \tfrac35x + y + \tfrac85z = x + y + z $
- $ \iff \tfrac35z - \tfrac25x = 0 $
- $ \iff \tfrac25x = \tfrac35z $
- $ \iff x = 1.5z $.
As both $ x, z \in \N $ (both have to be integers; there can be no "1.5 times 8 Coins"), the least possible solution is
- $ x = 3, z= 2 $.
This means that any number of 8-Coin results $ z $ where the number of 3-Coin results $ x = 1.5z $ is an integer delivers a number of Coins that can also be achieved by adding 5-Coin results the same number of times (see column 3a and 3b to the right for an example).
Other Coin combinations could basically be solved with the same equation system but would require more complex calculations if all $ a, b, c $ are involved.