a farmer is taking her eggs to market in her cart, but she hits a pothole, whick knocks over all the containers of eggs. Though she herself is unhurt, every egg is broken. So she goes to the insurance agent who asks her how many eggs she had. She says she doesnt know, but she remembers some things from various ways she tried packing the eggs. She knows that when she put the eggs in groups of two, there was one egg left over. When she out them in groups of three there was also one egg left over. The same thing happened when she put them in groups of four, groups of five, or groups of six. But when she put them in groups of seven, she ended up withcomplete groups of seven with no eggs left over. How many eggs did she have?

This is a particular case of a class of problems that can be solved by the Chinese remainder theorem.

In this particular case, it can be solved relatively easily without the theorem, since all the left-overs are one, except the last.

Since there was one egg left when packed in 2,3,4,5,6, we need to find the LCM (lowest common multiple) of 2,3,4,5,6 and add one to get the least possible answer, namely 60+1=61.
However, we recognize that 60k+1 will always satisfy the first 5 conditions, where k is a positive integer.

To satisfy the 7th condition, we need to find the least value of k such that 60k+1 is divisible by 7.

61 mod 7 = 5
121 mod 7 = 2
181 mod 7 = 6
241 mod 7 = 3
301 mod 7 = 0

So 301 is the least number of eggs the farmer had. She could have had 721 eggs (=301+420), but with the proof that she gave her insurance, she is likely to get paid for 301!

you have to find the least common multiple of 2, 3, 4, 5 and 6.

thus, LCM = 60
now, add 1 to 60, and check if the new number is divisible by 7,,
60 + 1 = 61
61/7 = 8 remainder 5
61 is not divisible by 7,, thus repeat this process using multiples of 60,, try 120,,
121/7 = 17 remainder 2

*after trial and error, i found 301,, just check if there is a number less than this which satisfies the conditions given.

so there,, =)

To find out the number of eggs the farmer had, we can use a technique called the Chinese Remainder Theorem.

Let's begin by examining the problem step by step.

1. The farmer puts the eggs in groups of 2, and 1 egg is left over. This means that the total number of eggs must be 1 more than a multiple of 2.

2. The farmer puts the eggs in groups of 3, and 1 egg is left over. This means that the total number of eggs must be 1 more than a multiple of 3.

3. The same applies when the eggs are grouped into 4, 5, and 6. In each case, there is always 1 egg left over.

4. However, when the eggs are grouped into 7, there are no eggs left over, meaning the total number of eggs is a multiple of 7.

Based on the information provided, we need to find a number that meets all these conditions.

Let's work through the process step by step:

Step 1: Find a multiple of 2, add 1, and check if it is divisible by 3, 4, 5, and 6.
- A multiple of 2 + 1 = 3. Not divisible by 3, 4, 5, and 6.

Step 2: Find the next multiple of 2, add 1, and check if it is divisible by 3, 4, 5, and 6.
- A multiple of 2 (2 x 2) + 1 = 5. Not divisible by 3, 4, 5, and 6.

Step 3: Find the next multiple of 2, add 1, and check if it is divisible by 3, 4, 5, and 6.
- A multiple of 2 (3 x 2) + 1 = 7. Divisible by 7.

So, the total number of eggs the farmer had is 7.

Therefore, the farmer had 7 eggs in her cart.

To solve this problem, we can use the concept of finding the least common multiple (LCM) of numbers. Here's how we can approach it:

Let's start by analyzing the information we have:

1. When the eggs are arranged in groups of 2, 1 egg is left over.
2. When the eggs are arranged in groups of 3, 1 egg is left over.
3. When the eggs are arranged in groups of 4, 1 egg is left over.
4. When the eggs are arranged in groups of 5, 1 egg is left over.
5. When the eggs are arranged in groups of 6, 1 egg is left over.
6. When the eggs are arranged in groups of 7, there are no eggs left over.

From this information, we can deduce that the number of eggs must be a multiple of 2, 3, 4, 5, and 6, but not of 7.

To find the least common multiple (LCM) of 2, 3, 4, 5, and 6, we can list out their multiples until we find a common multiple:

Multiples of 2: 2, 4, 6, 8, 10, 12, ...
Multiples of 3: 3, 6, 9, 12, ...
Multiples of 4: 4, 8, 12, 16, ...
Multiples of 5: 5, 10, 15, 20, ...
Multiples of 6: 6, 12, 18, ...

Based on this, we can see that 12 is the smallest number that is divisible by 2, 3, 4, 5, and 6.

Now let's check divisibility by 7. Starting from 12 and going in increments of 12, let's see when we find a number that is divisible by 7.

12, 24, 36, 48, 60, 72, 84, 96, ...

Here, we see that 84 is divisible by 7. So, the answer to the puzzle is that the farmer had 84 eggs in her cart.

Therefore, when packing the eggs into groups of 7, she ended up with exactly 12 groups of 7 eggs, with none left over.