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?

xXxTenTacion Jul 21, 2019

#2**+1 **

\(\text{Well let's turn this into a couple equations if nothing else}\\ N = 7j \\ N = 0 \pmod{7}\\ N+53 = 8k\\ N+56-3=8k\\ N-3=8(k-7)\\ N=3 \pmod{8}\)

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

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

.Rom Jul 21, 2019

#3