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How many ordered pairs (x,y) of positive integers satisfy the inequality 4x + 5y < 200?

 Jun 28, 2021
 #1
avatar+151 
+2

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}  $

 Jun 28, 2021
edited by UsernameTooShort  Jun 28, 2021
 #2
avatar+121062 
+2

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

 

cool cool cool

 Jun 28, 2021
edited by CPhill  Jun 28, 2021

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