# Problem of the Week

Topics: Division, Quotient, Modular arithmetic Pages: 2 (587 words) Published: October 25, 2012
Shania Lewis
Period 6B
September 4, 2012
Problem of the Week: The Broken Eggs

Problem Statement: A farmer is carrying her eggs in a cart when she accidentally spills every one of them and they all break. She decides to go to her insurance agent, who asks her how many eggs she had. She’s not sure but she does know some information from various ways she tried to pack her eggs. She knows that when she put her eggs in groups of one, two, three, four, five, and six, there was always one egg left over. When she put them in groups of seven, there were no eggs left over. The task is to use this information to find out how many eggs the farmer had,

Process: In the process of attempting to solve this problem, I first figured out that I would have to find a number that leaves a remainder of 1 when divided by 2, 3, 4, 5, and 6 but is evenly divisible by 7. I noticed I only need to look at numbers that were a multiple of 7. 7/2= 3.1 7/3=2.1 7/4=1.3 7/5=1.2 7/6=1.1

When I divided all the numbers by 7 I saw that 2, 3, and 6 all leave a remainder of 1, but not 4 and 5. 14/2=7 14/3=4.2 14/4=3.2 14/5=2.4 14/6=2.2
I tried this again with the number 14 since it was a multiple of 7, but then I realized this would be too long of a process to do this with a different number over and over again. I thought about what I was doing and I knew that an even number wouldn’t work because they are divisible by 2. I realized that the number has to be an odd number and end with 1. I started to only look at numbers that ended with 1. 91 161 231

91 didn’t because 4 didn’t have a remainder of 1. 161 didn’t work because 2 didn’t have a remainder of 1. 231 didn’t work because 2 didn’t have a remainder of 1. Then I got to 301, and 2, 3, 4, 5, and 6 all had a remainder of 1 when I divided them.

Solution: My solution to this problem is the farmer had 301 eggs. This solution is correct because 301 is a multiple of 7, and which it is divided by 2, 3, 4, 5, and 6 it leaves a remainder of 1. I...