jmc

Question

You have seven bags of gold coins. Each bag has the same number of gold coins. One day, you find a bag of 53 coins. You decide to redistribute the number of coins you have so that all eight bags you hold have the same number of coins. You successfully manage to redistribute all the coins, and you also note that you have more than 200 coins. What is the smallest number of coins you could have had before finding the bag of 53 coins?

Accepted Answer

If there are b gold coins in each of the original 7 bags, then 7b+53 is divisible by 8. In other words, 7b + 53 0 8. Since 53 5 8 and 7 -1 8, we have that -b -5 8. Multiplying both sides by -1, we get that b 5 8. Now, we want 7b + 53 > 200, so as a result, b > 200-537 b > 21. Therefore, we want an integer greater than 21 which leaves a remainder of 5 when divided by 8. The least such integer is 29, so you had 29 7 = 203 coins before finding the bag of 53 coins.