1) The number of ordered paurs of positive integer (m,n), such that 1/n + 1/m = 1/15
2) Consider the equation 2u + v + w + x + y + z = 3. How many solutions (u , v , w , x , y , z) of non-integers does this equation have?
+655 1) We can rewrite this as (n + m)/mn = 1/15, which also is mn = 15m + 15n.
We move everything to one side:
mn - 15m - 15n = 0
There is this method of adding a special number to both sides to make this doable.
In this case, we do + ( - 225) to both sides. mn - 15m - 15n - 225 = -225
We can rewrite this as -n*(m-15)-15(m-15)=-225, which also is (m-15)*(n-15) = 225
Using a factoring trick, 225 can be written as (1,225);(3,75);(5,45);(9,25);(15,15)
So there are nine ordered pairs because 15,15 is the same as 15,15. Others can be doubled.
