document.write( "Question 1099298: There are 100 seats for 100 people at the movie theater. The total price for the 100 people is $1.00. Men will be charged $.05 each, women $.02 and children 10 for $.01. How many men, women, and children will there be? \n" ); document.write( "
Algebra.Com's Answer #713732 by greenestamps(13200)\"\" \"About 
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Let m be the number of men, w the number of women, and c be the number of children. Then

\n" ); document.write( "\"m%2Bw%2Bc+=+100\" [the total number of people is 100]
\n" ); document.write( "\".05m%2B.02w%2B.001c+=+1\" [the total cost of the tickets is $1.00]

\n" ); document.write( "This is a problem with Diophantine equations. Diophantine equations are systems of equations in which there are fewer equations than variables, but in which it is possible to find the solution (or a family of solutions) using the requirement that all the variables must have values that are integers (usually, as in this case, positive integers).

\n" ); document.write( "With two variables and three equations, the procedure is to use the two equations to eliminate one of the variables, leaving one equation with two variables.
\n" ); document.write( "Then solve that one equation for one variable in terms of the other, and use the requirement that both variables have positive integer values to find solution(s).

\n" ); document.write( "With the two equations in this problem, we can eliminate variable c by multiplying the second equation by 1000 and subtracting the two equations:

\n" ); document.write( "\"50m%2B20w%2Bc+=+1000\"
\n" ); document.write( "\"m%2Bw%2Bc+=+100\"
\n" ); document.write( "\"49m%2B19w+=+900\"
\n" ); document.write( "\"19w+=+900-49m\"
\n" ); document.write( "\"w+=+%28900-49m%29%2F19\"

\n" ); document.write( "With the equation in this form, we know that, because w has to be an integer, (900-49m) must be divisible by 19.

\n" ); document.write( "There are various ways to find what value(s) of m make this an integer. With pencil and paper, the best way is to rewrite the numerator (900-49m) in two parts, the first of which is exactly divisible by 19.

\n" ); document.write( "Proceeding with that idea...
\n" ); document.write( "\"w+=+%28912-57m%29%2F19+%2B+%288m-12%29%2F19+=+%2848-3m%29+%2B+4%282m-3%29%2F19\"

\n" ); document.write( "With the equation in this form, we see that (2m-3) must be divisible by 19; and some quick calculations show that m must be 11. Then
\n" ); document.write( "\"w+=+%2848-33%29+%2B+4%2819%2F19%29+=+15%2B4+=+19\"

\n" ); document.write( "And then since there are 11 men and 19 women, the number of children must be 70.


\n" ); document.write( "Now here is a different approach to the problem which uses more standard algebraic techniques.

\n" ); document.write( "Since the price for children is 10 for a penny, the number of children must be a multiple of 10.
\n" ); document.write( "Furthermore, a bit of logical analysis tells us that, since that average cost of the tickets is a penny, most of the people must be children.

\n" ); document.write( "So see if you can find a solution if the number of children is 90; if not, then try 80; then 70; and so on, until you find a solution.

\n" ); document.write( "If c = 90...
\n" ); document.write( "\"m%2Bw+=+10\"
\n" ); document.write( "\"5m%2B2w+=+91\"
\n" ); document.write( "Multiply the first equation by 5 and then subtract the two equations:
\n" ); document.write( "\"5m%2B5w+=+50\"
\n" ); document.write( "\"3w+=+-41\" obviously no good...

\n" ); document.write( "If c = 80...
\n" ); document.write( "\"m%2Bw+=+20\"
\n" ); document.write( "\"5m%2B2w+=+92\"
\n" ); document.write( "\"5m%2B5w+=+100\"
\n" ); document.write( "\"3w+=+8\" no; 8 is not divisible by 3

\n" ); document.write( "If c = 70...
\n" ); document.write( "\"m%2Bw+=+30\"
\n" ); document.write( "\"5m%2B2w+=+93\"
\n" ); document.write( "\"5m%2B5w+=+150\"
\n" ); document.write( "\"3w+=+57\"
\n" ); document.write( "\"w+=+19\"
\n" ); document.write( "This works. c = 70; w = 19; that makes m = 11.
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