Word Problems/Gold Coins

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GGG wrote at 2014-07-19 07:48:21
It is easiest to start at the end, and work your way to the beginning.



When the pirates wake up in the morning, there must be 3a + 1 coins, where a is any non-negative integer. They divide it 3 ways with one left over.



When the third pirate wakes up, he finds 3/2 (3a + 1) + 1 = 9/2 * a + 3/2 + 1 coins.



When the second pirate wakes up, she finds 3/2 (9/2 * a + 3/2 + 1) + 1 = 27/4 * a + 9/4 + 3/2 + 1 coins.



When the first pirate wakes up, he finds 3/2 (27/4 * a + 9/4 + 3/2 + 1) + 1 = 81/8 * a + 27/8 + 9/4 + 3/2 + 1 coins. This is the starting amount. 81/8 * a + 27/8 + 9/4 + 3/2 + 1 = 81/8 * a + 27/8 + 18/8 + 12/8 + 8/8 = 81/8 * a + 65/8 = (81a + 65) / 8



The starting amount of coins, (81a + 65) / 8, must be a whole number. That means that 81a + 65 must be divisible by 8. If a = 0, the remainder will be 1. If a = 1, the remainder will be 2. All the way up until a = 7, and the remainder is 0. So a = 8k + 7, where k is any non-negative integer. (81(8k + 7) + 65) / 8 = (81 * 8k + 81*7 + 65) / 8 = (81 * 8k + 632) / 8 = 81k + 79



The fewest amount of coins the pirates could start with is where k = 0, so the number of coins = 81*0 + 79 = 79 coins.



They could also start with 79 + any positive multiple of 81.


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Stephen King

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I can help with all word problems for any math (including college math). I can show how to set up the problem, and how I came about that solution. Some word problems rely on many facets of math to solve, and it`s important for someone to realize that.

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