Posted by Peter on Friday, September 30, 2011 at 1:58am.
number of smaller size : x
number of larger size : y
33x + 103y = 6526
There are several ways to solve these for positive integer solutions.
I use a method involving "continued fractions" , which I will not even attempt to show here.
With my method , I found that
x = 101 and y = 31 gives a solution
Here is a page which is very similar to what I do
http://www.wikihow.com/Solve-a-Linear-Diophantine-Equation
Professor Burris from the University of Waterloo (my old school) has written hundreds of little algorithms to solve these kind of problems.
here is one of them for our problem
http://www.math.uwaterloo.ca/~snburris/htdocs/linear.html
enter :
33 for a, 103 for b, and 6526 for c
This will give us an initial solution of
x = 163150 and y = -52208
Of course this is no good, since both our values must be positive, but ....
the slope of our linear equation is -33/103
which means that for every increase of 33 in -52208 we have to decrease 163150 by 103
so I increased and decreased by a factor of 1583
x = 163150 - 1583(103) = 101
y = -52208 + 1583(33) = 31
btw, I got the 1583 by using the lower integer value of 163150/103
By successive reductions:
1--33x + 103y = 6526
2--Dividing by the lowest coefficient
---x + 3y + 4y/33 = 197 + 25/33
---(4y - 25)/33 = 197 - x - 3y
3--(4y - 25)/33 must be an integer.
4--Multiplying (3) by 25 and redividing
---(100y - 625)/33 =
---3y + y/33 -18 - 31/33
5--(y - 31)/33 must be an integer k ---making y = 33x + 31
6--Substituting back into (1) yields
---x = 101 - 103k
7--By inspection, the only possible ---value of k is zero making
---x = x = 101 and y = 31
8--Checking: 33(101) + 103(31) =
---3333 + 3193 = 6526
The Euclidian Algorithm is another approach to problems of this sort.