Imp Pow 1: the Broken Eggs
IMP POW 1: The Broken Eggs
Problem Statement:
A farmer’s cart hits a pothole, causing all her eggs to fall out and break. Luckily, she is unhurt. To cover the cost of the eggs, her insurance agent needs to know how many she had. She can’t remember the number, but can remember some problems she had when packing the eggs. When she put the eggs in groups of two to six eggs, there was always one left over. However, in groups of seven, there were none left over. From what she knows, how can she figure out how many eggs she had?
Process:
First, I thought the answer would be forty-nine. Then, I realized my mistake and tried to think of different ways to do it. I decided to make a chart showing the remainders of the numbers two to six into multiples of seven. I then started to find a pattern. I noticed that the number two has a remainder of one every other multiple of seven, the number three has a remainder of one every two multiples, the number four has a remainder of one every three multiples, and so on. I marked a dot every time there was a remainder of one. I knew that when I had dots marked from two through seven, that would be the answer. So I set out on the long journey of calculating when there was a remainder of one until I reached a number that was all filled up.
Solution:
Using my process, I found that the number of eggs the farmer had was three-hundred one. I know this because it was the first number to fit my boundaries. This number, when put in groups of two to six, will have a remainder of one, but when put in groups of seven, will magically have no remainder. I can assume that there would be another possible larger number that would work, but unfortunately I was too lazy efficient to calculate it.
Evaluation:
Trying to figure out how to solve this problem was very hard. However, once I knew how, it was easier. I think this problem definitely benefited me. I learned that no matter how hard a problem looks. You can always almost solve it. I
Problem Statement:
A farmer’s cart hits a pothole, causing all her eggs to fall out and break. Luckily, she is unhurt. To cover the cost of the eggs, her insurance agent needs to know how many she had. She can’t remember the number, but can remember some problems she had when packing the eggs. When she put the eggs in groups of two to six eggs, there was always one left over. However, in groups of seven, there were none left over. From what she knows, how can she figure out how many eggs she had?
Process:
First, I thought the answer would be forty-nine. Then, I realized my mistake and tried to think of different ways to do it. I decided to make a chart showing the remainders of the numbers two to six into multiples of seven. I then started to find a pattern. I noticed that the number two has a remainder of one every other multiple of seven, the number three has a remainder of one every two multiples, the number four has a remainder of one every three multiples, and so on. I marked a dot every time there was a remainder of one. I knew that when I had dots marked from two through seven, that would be the answer. So I set out on the long journey of calculating when there was a remainder of one until I reached a number that was all filled up.
Solution:
Using my process, I found that the number of eggs the farmer had was three-hundred one. I know this because it was the first number to fit my boundaries. This number, when put in groups of two to six, will have a remainder of one, but when put in groups of seven, will magically have no remainder. I can assume that there would be another possible larger number that would work, but unfortunately I was too lazy efficient to calculate it.
Evaluation:
Trying to figure out how to solve this problem was very hard. However, once I knew how, it was easier. I think this problem definitely benefited me. I learned that no matter how hard a problem looks. You can always almost solve it. I