Question 1195792: 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
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52803)   (Show Source):
Answer by greenestamps(13200)  (Show Source):
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Let x be the number of coins in each of the original 7 bags; the total number of coins originally was then 7x.
After finding an 8th bag containing 53 coins, the total number of coins is 7x+53.
That total number of coins can be distributed equally into 8 bags, so 7x+53 is a multiple of 8. So we need to find solutions in positive integers of the equation
This is a linear Diophantine equation -- a single equation with two unknowns, whose solution(s) can be found knowing that both variables have integer values.
One standard method for finding the solutions is to solve the equation for one variable in terms of the other, as follows.
[1]
 [2]
 [3]
In that equation, x is a positive integer; and y has to be a positive integer. That means (53-x)/8 is an integer.
Note that we could have looked for solutions in positive integers to equation [1] itself; however, performing the steps in [2] and [3] gives us an equation for which it is much easier to find the solutions.
Remember that we are trying to find the smallest value of 7x, which is the original number of coins, given that the final total number of coins, 7x+53, is greater than 200.
x y 7x 7x+53
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5 5+48/8=11 35 88
13 13+40/8=18 91 144
21 21+32/8=25 147 200
29 29+24/8=32 203 256
The smallest value of 7x for which 7x+53 is greater than 200 is 203.
ANSWER: The total number of coins before finding the 8th bag was 203.
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