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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?

tertre  Mar 12, 2018
 #1
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7 bags  each have 29 coins to start    .

 

Let x be the final coin coint in the EIGHT bags     8x > 200   so x >25

 

Now    53 - x    has to be divisible by the remaining 7 bags

trial and error 

26     53 - 26 = 27   nope

27     53 - 27 = 26   nope

.

.

.

.

32    53-32 = 21   Yes !       x = 32 = final count in the 8 bags

 

(8 * 32   -53  ) / 7 = inital coin count in the 7 bags = 29 coins/bag

ElectricPavlov  Mar 12, 2018
edited by ElectricPavlov  Mar 12, 2018

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