Suppose she has n eggs.  Then n-1 is both a multiple of 3 and
5 which means that n-1 is a multiple of 15. if she removes the 
eggs 7 at a time, there are no eggs left. That means n is a 
multiple of 7.

Since n-1 is a multiple of 15, there exists positive integer p 
so that n-1 = 15p

Since n is a multiple of 7, there exists positive integer q 
so that n = 7q.

So we have this system of equations:



So we substitute 7q for n from the second equation, into the first equation







The smallest coefficient is 7, and it has absolute value 7,
so write all numbers in terms of their nearest multiple of 7.





Divide every term through by 7





Isolate the fraction terms:



The left side is an integer and so is the right side.
Let that integer be A

then we have this system:
 


Clear the second of fractions by multiplying through by 7



Solve the second one for p



Substitute 7A-1 for p in the other equation









Substitute that in



















Therefore A = 1 because 1 is the only integer
between those two values.

 Since 










  
 



Therefore she has 91 eggs in her basket.

A woman with a basket of 91 eggs finds that if she removes
the eggs from the basket 3 at a time as we can see from the
first two divisions below that after she removes 30 groups 
of 3 or 18 groups of 5 she will in both cases have 1 remaining 
because there is a 1 remainder in the first two cases.  The 
third division below shows that when she removes 13 groups of 7 
each she has none remaining because there is a 0 remainder.

  30         18       13
3)91       5)91     7)91
  90         5        7
   1         41       21
             40       21
              1        0

Edwin