Question 357851
 A woman with a basket of eggs finds that if she removes
the eggs from the basket 3 or 5 at a time, there is always 1
egg left. However, if she removes the eggs 7 at a time,
there are no eggs left. If the basket holds up to 100 eggs,
how many eggs does she have? Explain your reasoning.
<pre>
Suppose she has N eggs

>>...the eggs from the basket 3 or 5 at a time, there is always 1
egg left...<<

That means that N is 1 more than a multiple of both 3 and 5, which means
that it is 1 more than a multiple of 15.  So we can write N as 15p + 1, where k
is a positive integer.

>>...However, if she removes the eggs 7 at a time, there are no eggs left...<<

So we have to find a positive integer of the form 15p + 1 that is a multiple of 7.

Try p=1.  15p + 1 = 15(1) + 1 = 16, not a multiple of 7.
Try p=2.  15p + 1 = 15(2) + 1 = 31, not a multiple of 7.
Try p=3.  15p + 1 = 15(3) + 1 = 46, not a multiple of 7.
Try p=4.  15p + 1 = 15(4) + 1 = 61, not a multiple of 7.
Try p=5.  15p + 1 = 15(5) + 1 = 76, not a multiple of 7.
Try p=6.  15p + 1 = 15(6) + 1 = 91, which is a multiple of 7, so that's the answer.

91 eggs.  You can also do it by solving the Diopantine equation

                  15p + 1 = 7q  

Are you supposed to do it that way?  That would depend on what 
course of mathematics you are taking.

91 = 30*3+1 = 28*5+1 = 13*7

So 

when she takes out 30 groups of 3 eggs each, she takes out 90 eggs, and has 1 egg left,
when she takes out 18 groups of 5 eggs each, she takes out 90 eggs, and has 1 egg left, 
when she takes out 13 groups of 7 eggs each, she takes out all 91 eggs and has none left.

Edwin</pre>