Question 331715
 Hi there, I am stumped at how to do about this. Can someone please help me? Thank you! 
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><b> 
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:

{{{system(n-1=15p,n=7q)}}}

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

{{{n-1=15p}}}

{{{7q-1=15p}}}

{{{7q-15p=1}}}

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

{{{7q-(14+1)p=0+1}}}

{{{7q-14p-p=1}}}

Divide every term through by 7

{{{7q/7-14p/7-p/7=1/7}}}

{{{q-2p-p/7=1/7}}}

Isolate the fraction terms:

{{{q-2p=p/7+1/7}}}

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

then we have this system:
 
{{{system(q-2p=A, p/7+1/7=A)}}}

Clear the second of fractions by multiplying through by 7

{{{system(q-2p=A, p+1=7A)}}}

Solve the second one for p

{{{p=7A-1}}}

Substitute 7A-1 for p in the other equation

{{{q-2p=A}}}

{{{q-2(7A-1)=A}}}

{{{q-14A+2=A}}}

{{{q=15A-2}}}

Substitute that in

{{{n = 7q}}}

{{{n = 7(15A-2)}}}

{{{n = 105A-14}}}

{{{0<n <= 100}}}

{{{0<105A-14 <=100}}}

{{{14< 105A <=114}}}

{{{14/105<A <= 114/105}}}

{{{2/15<A <= 38/35}}}

{{{2/15<A <= 1&3/35}}}

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

 Since 

{{{p=7A-1}}}
{{{p=7(1)-1}}}
{{{p=7-1}}}
{{{p=6}}}

{{{q=15A-2}}}
{{{q=15(1)-2}}}
{{{q=15-2}}}
{{{q=13}}}
  
{{{n = 7q}}} 
{{{n = 7(13)}}}
{{{n =  91}}}

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.

 <u> 30</u>        <u> 18</u>      <u> 13</u>
3)91       5)91     7)91
  <u>90</u>         <u>5</u>        <u>7</u>
   1         41       21
             <u>40</u>       <u>21</u>
              1        0

Edwin</pre>