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

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 all the eggs are 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 some things from various ways she tried paying the eggs. When she put the eggs in groups of two,three,four,five and six there was always one egg leftover, but when she put them in groups of seven they ended up in complete groups with no eggs left over .

 

 

Answer 1

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.

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 w/div by 7: 5     2     6    3     0     4    1     5    2 (REPEATS!)

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.)