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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 500 coins. What is the smallest number of coins you could have had before finding the bag of 53 coins?

 Nov 6, 2020
 #1
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536  coins all together

Start with 7 bags of 69 each =483 coins to start

 Nov 6, 2020
 #2
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But how did you GET that answer? 

 

Guest Nov 6, 2020
 #3
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(7*n + 53 ) mod 8=0.........................(1) 

7*n + 53 > 500, solve for n...............(2)

 

n =8m + 5    and n >=8, where m =0, 1, 2, 3........etc.

Therefore, m = 8 and:

8 x 8 + 5 =69 - original number of gold coins in each of the 7 bags.

69 x 7 =483 - total of all original coins.

[69 - 2] =67 - coins in each of the 7 original bags after removing 2 coins from each bag.

[53 + (7 x 2) ] =53 + 14 =67 gold coins in the 53-coin bag after removing 2 coins from each of the original 7 bags.

8 x 67 =536 coins after balancing the 8 bags with 67 coins in each bag.

 Nov 6, 2020

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