HELP ME PLEASE

Question

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? Is there more than one answer?

Answer 1

The problem gives you a great opportunity to learn about the Least Common Multiple (LCM) of numbers.  For example, if you have the numbers 3 and 4, the multiples are:

  of 3:     3   6   9  12  15  18  21  24  27  30

  of 4:     4   8  12  16  20  24  28  32  36

The least (lowest) common (in both lists) multiple (the LCM) is 12.

 

The LCM may also be found by selecting just enough of the prime factors of each number such that the LCM is least.  For example, find the LCM of 12 and 8:

prime factors of 12:   2*2*3

prime factors of 8:    2*2*2

LCM = 2*2*2*3 = 24   [both 8 and 12 can be made from these facors]

 

For this problem,

 

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 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.)  [note: 420 is the LCM of 2,3,4,5,6,7]

Answer 2

Using the first clue, we know that the number must be odd because it is not divisible by an even number.

We also know that the number is an uneven multiple of 7 (7x1, 7x3, 7x5, etc)

 

I would say that she could have 7, 21, 35, 49, ... eggs.