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November 24, 2014

November 24, 2014

Posted by **Peter** on Friday, September 30, 2011 at 1:58am.

- Diophantine Equations (Algebra 2) -
**Reiny**, Friday, September 30, 2011 at 8:53amnumber 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

- Diophantine Equations (Algebra 2) -
**tchrwill**, Friday, September 30, 2011 at 1:05pmBy 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.

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