How many ordered pairs (x,y) of positive integers satisfy the inequality 4x + 5y < 200?
+171 what you are given is $ \{ 4x + 5y < 200 \} $
now, how many pairs of coordinates are there to satisfy that, and also $ ∈ \mathbb{Z} $?
a graph $ \{ 4x + 5y = 200 \} $ -- you will see a line -- the coordinates we wont need are the ones which are larger than 200, so we just expunge the top part, where the number gets higher right?
(NOTE: 0 is not a positive number)
check the image
do not forget to include the coordinates where the purple line is on as well
at this point just count them all if you want an answer -- unless someones manage to find an easier way LOL
also, do not count the ones where the line intersects, as those would make it $=200$ $ \overset{. \: .}{\smile} $
+129850 As UTS said, we can solve this geometrically using something known as Pick's Theorem
If we graph 4x + 5y = 200........this will form a first quadrant triangle with a base of 50 and a height of 40
The area = (1/2)(40(50) = 1000
Pick's Theorem says that the area =
Number of lattice points in the interior of the triangle + number of lattice points on the boundary of the triangle / 2 - 1
Where a lattice point is a point with integer coordinates
The number of lattice points on the boundary of this triangle = 41 + 50 + 9 = 100
So
1000 = number of lattice points in the interior + 100/2 -1
1000 = number of lattice points in the interior + 49
1000 - 49 = number of latttice points in the interior
951 = number of lattice points in the interior = number of ordered pairs of positive integers satisfying
4x + 5y < 200
See the graph here : https://www.desmos.com/calculator/tkypy0yfbs
