+586 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?
+6250 \(\text{Well let's turn this into a couple equations if nothing else}\\ N = 7j \\ N = 0 \pmod{7}\\ N+53 = 8k\\ N+56-3=8k\\ N-3=8(k-7)\\ N=3 \pmod{8}\)
\(\text{So we have the system}\\ N=0 \pmod{7}\\ N=3 \pmod{8}\)
\(\text{Just by inspection $N=35$}\)
.+6250 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.
