Question 710238
<pre>
     m + w + c = 100
5m + 2w + .10c = 100

The number of children is a multiple of 10, so 
it will come out to a whole number of dollars.

Let c = 10k

    m +  w + 10k = 100
   5m + 2w +   k = 100

Since these both = 100, set them equal:

  m + w + 10k = 5m + 2w + k

      9k - 4m = w

Substitute 9k - 4m for w in the first equation

       m +  w + 10k = 100
m + (9k - 4m) + 10k = 100
  m + 9k - 4m + 10k = 100
          -3m + 19k = 100

the smallest coefficient in absolute value is 3
So we write 19 and 100 in terms of their nearest
multiple of 3. So we write 19 as 18+1 and 100 as 99+1

      -3m + (18+1)k = 99+1
      -3m + 18k + k = 99 + 1 

Divide through by 3

       -m + 6k + {{{k/3}}} = 33 + {{{1/3}}}

Isolate the fractions:

       {{{k/3}}} - {{{1/3}}} = 33 + m - 6k

The right side is an integer, and so is the left side,
so set both sides = A, an integer:

{{{k/3}}} - {{{1/3}}} = A;     33 + m - 6k = A
 k - 1 = 3A
     k = 3A+1

Substitute 3A+1 for k in     33 + m - 6k = A
                        33 + m - 6(3A+1) = A
                        33 + m - 18A - 6 = A 
                                       m = 19A - 27

Substitute 3A+1 for k  and 19A-27 for m in            9k - 4m = w 
                                          9(3A+1) - 4(19A-27) = w
                                          27A + 9 - 76A + 108 = w
                                                    117 - 49A = w

So we have  m = 19A-27 men, 117-49A = w women, and c = 10k = 10(3A+1) 
= 30A+10 children  

We can't have a negative number of men, women or children

        m &#8807; 0 
 19A - 27 &#8807; 0
      19A &#8807; 27
        A &#8807; {{{1&8/19}}}

        w &#8807; 0 
117 - 49A &#8807; 0
     -49A &#8807; -117
        A &#8806; {{{2&19/49}}}

Thus {{{1&8/19}}} &#8806; A &#8806; {{{2&19/49}}}

Therefore A = 2 since that is the only integer that
satisfies that inequality.

 m = 19A - 27 = 19(2) - 27 = 38 - 27 = 11 men
 w = 117 - 49A = 117 - 49(2) = 117 - 98 = 19 women
 c = 30A + 10 = 30(2) + 10 = 60 + 10 = 70 children
 
Answer:  11 men, 19 women, and 70 children.

Edwin</pre>