Maths Problem Using LCM

Question:

A farmer is taking her eggs to the market in a cart, but she hits a pothole, which knocks over all the containers of eggs. Though she is unhurt, every egg is broken. So she goes to her insurance agent, who asks her how many eggs she had. She says she doesn't know, but she remembers somethings from various ways she tried packing the eggs.

When she put the eggs in groups of two, three, four, five, and six there was one egg left over, but when she put them in groups of seven they ended up in complete groups with no eggs left over.

What can the farmer figure from this information about the number of
eggs she had?

Answer:

Let N = total number of eggs.

N is not evenly divisible by 2, 3, 4, 5, or 6. There is always 1 left over.
N is evenly divisible by 7. There is 0 left over.

So, we want to find an N that is a multiple of 7 and a (N-1) that is a common multiple of 2, 3, 4, 5, and 6.

For 2, 3, 4, 5, and 6 the prime factors are:
2:   2
3:   3
4:   2*2
5:   5
6:   2*3
So, the LCM is 2*2*3*5 = 60.

Find the multiples of 60 and add 1 (since dividing by 2, 3, 4, 5, and 6 always has a remainder of 1).

The multiples of that LCM are:
 60  120  180  240  300  360  420  480  540

                     

Add 1:  

61  121  181  241  301  361  421  481  541
 

Remainder when divided by 7:    

 5      2     6      3      0     4      1     5      2   

The remainder of (N divided by 7) is 0 (it is evenly divisible). So, N=301.

(Note: N could also be 721, 1141, 1561, 1981, 2401, 2821, 3241, 3661, etc., but 301 is the least value. Since the remainder repeats, you can find each of these values by adding 420.) 

[Note: 420 is the LCM of 2,3,4,5,6,7]