A group of bandits managed to rob a train. They absconded with a bag of gold pieces.
When they camped for the night, they took turns watching for the posse. Alan took first watch. While setting there in the dark, looking at the bag of gold… he got to wondering what their take was. He knew that a full bag held 5000 gold coins, and this bag was obviously not full. Alan set down and divided the gold into piles for each of the bandits. When the bag was empty he found that he had one coin left over. Knowing how argumentative his colleagues are - he decided to alleviate the argument by throwing the extra coin into the river. Alan then put his share of the coins in his saddlebag and returned the rest to the bag.
Bob took the second watch. In order to keep himself awake, Bob decided to split up the shares. He took the coins from the bag and separated them into even piles - one for each bandit. When the bag was empty, he had one coin left. Rather than risk an argument - he threw the extra coin into the river. Bob put his share in his saddlebags and returned the rest of the coins to the bag.
Carl had next watch. He also decided to divide the loot. He also found that once the coins were divided evenly, there was an extra coin. He tossed it into the river, put his share in his saddlebags, and returned the rest to the bag.
Dan had the next watch. Dan divided the loot, found an extra coin, threw it into the river, put his share in his saddlebags and returned the rest to the bag.
Ernie was the last bandit and had the final watch. He divided the coins into piles for each bandit and was pleased to find that they divided evenly.
How many coins were in the bag to start with?
The bag had to be empty prior to filling any gold pieces into it whether filled by the robbers or filled at the bank or wherever it was filled with the gold. All bags are empty before you put something into it.
{([{(5x)5/4+1}5/4+1]5/4+1)5/4+1}=y or 3125/256x+369/64=y
trial and error here..
knew ernie had to have a set amount of coins that dived equally by 5 so that meant 5,10,15,20,25,etc all the way up to about 2000, much higher than that and it meant there must of been more than 5000 coins in the bag to start with.
so then dived the start number by 4 and multiplied by 5(the start number had to be 4/5ths of what the last guard had), if i got an whole number answer added 1(the coin that was thrown away) coin to it and repeated, divided by 4 x 5,
kept doing it until i either got an answer that wasn't whole, meant it was the wrong start number, or i kept getting whole number answers for the 4 repeats needed to get the correct answer needed.
tried seeing if i could see a pattern in my results that would make it easier to get the answer but nothing i could pin down, except only needed to use start numbers ending in 00,20,40,60,80. so cut down the numbers i needed to try considerably, still took me an hour or more to get to the magic start number of 1020.
ernie 1020/4x5=1275
dan 1276/4x5=1595
carl 1596/4x5=1995
bob 1996/4x5=2495
alan started with 2496
{([{(5x)5/4+1}5/4+1]5/4+1)5/4+1}=y or 3125/256x+369/64=y
trial and error here..
knew ernie had to have a set amount of coins that dived equally by 5 so that meant 5,10,15,20,25,etc all the way up to about 2000, much higher than that and it meant there must of been more than 5000 coins in the bag to start with.
so then dived the start number by 4 and multiplied by 5(the start number had to be 4/5ths of what the last guard had), if i got an whole number answer added 1(the coin that was thrown away) coin to it and repeated, divided by 4 x 5,
kept doing it until i either got an answer that wasn't whole, meant it was the wrong start number, or i kept getting whole number answers for the 4 repeats needed to get the correct answer needed.
tried seeing if i could see a pattern in my results that would make it easier to get the answer but nothing i could pin down, except only needed to use start numbers ending in 00,20,40,60,80. so cut down the numbers i needed to try considerably, still took me an hour or more to get to the magic start number of 1020.
ernie 1020/4x5=1275
dan 1276/4x5=1595
carl 1596/4x5=1995
bob 1996/4x5=2495
alan started with 2496
That's the same thing the formula says… with the exception that the formula starts with the number of coins in each pile when Ernie separates them… so the magic x is 204 - which means there was 1020 coins in the bag when he received it. Still takes trial and error… but I was too lazy to do it that way so I just plugged the formula into excel and had it run all the iterations of x up to 500. The first whole number result for y was 2496 where x=204
too much for me, whew.
trial and error here.. 
knew ernie had to have a set amount of coins that dived equally by 5 so that meant 5,10,15,20,25,etc all the way up to about 2000, much higher than that and it meant there must of been more than 5000 coins in the bag to start with.
so then dived the start number by 4 and multiplied by 5(the start number had to be 4/5ths of what the last guard had), if i got an whole number answer added 1(the coin that was thrown away) coin to it and repeated, divided by 4 x 5,
kept doing it until i either got an answer that wasn't whole, meant it was the wrong start number, or i kept getting whole number answers for the 4 repeats needed to get the correct answer needed.
tried seeing if i could see a pattern in my results that would make it easier to get the answer but nothing i could pin down, except only needed to use start numbers ending in 00,20,40,60,80. so cut down the numbers i needed to try considerably, still took me an hour or more to get to the magic start number of 1020. 
