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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?

 Jul 21, 2019
 #1
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 Jul 21, 2019
edited by Guest  Jul 21, 2019
 #2
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\(\text{Well let's turn this into a couple equations if nothing else}\\ N = 7j \\ N = 0 \pmod{7}\\ N+53 = 8k\\ N+56-3=8k\\ N-3=8(k-7)\\ N=3 \pmod{8}\)

 

\(\text{So we have the system}\\ N=0 \pmod{7}\\ N=3 \pmod{8}\)

 

\(\text{Just by inspection $N=35$}\)

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 Jul 21, 2019
 #3
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Rom: The question says " You have more than 200 coins AFTER the  8th bag of 53"

 Jul 21, 2019
 #4
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roger

 

7 and 8 are relatively prime so the solutions will be separated by their product, i.e. 56.

35 + 2 x 56 = 147

 

so adding 53 gets us exactly 200, so it must be the next one

 

35+3x56 = 203

 

is the first solution that meets the requirements.

 

The original number of coins was 203.

Rom  Jul 21, 2019

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