Find the number positive integer solutions of 4x+6y=200
\(\large \color{black}{\begin{align} & \normalsize \text{Find the number positive integer solutions of}\hspace{.33em}\\~\\ & 4x+6y=200 \end{align}}\)
I need to find quick way
I don't really know but you should look into the Euclidean Algorithm I think that might be a good path to start looking into.
as this is Diophantine equation i agree with @Empty
so lets try 2x+3y=100
yeah both equations are same but we don't really need euclid algorithm here as finding a particular solution seems kinda easy by inspection. By inspection, (50, 0) is one particular solution
but Euclid would help in finding general solution to show it's infinite solutions right ?
euclid is for finding particular solution since we already figured that (50, 0) is a particular soltion, we can avoid euclid
:3
I found out (50,0),(47,2),.....(2,32) but that took more than 2 min
finding "one" particular solution is enough
anyway i like this way to show its infinite:- |dw:1442748565442:dw|
after you have one particular solution, try finding the "null" solution : 2x + 3y = 0 Easy to see that (-3, 2) solves above equation. Therefore the complete solution is given by : `(50, 0) + t(-3, 2)`
since you want just the positive integer solutions, solve : 50 - 3t > 0 0 + 2t > 0
50/3=16.
answer is 17
right, solve it simultaneously you should get an interval of "t" as solution
aha i haven't note positive :O
I think the answer should be 16
0<t<16
nope 0 < t < 16.66 there are exactly 16 positive integers in that interval
0<t<17
0 < t < 50/3 leave it like that
ok u r right (50,0) doesn't count answer is 16
Yep! lets do one more example maybe ?
Find the number of positive integer solutions to the equation 7x + 13y = 700
If you prefer, here are the steps : 1) Find any one `particular` solution by inspection 2) Find the `null` solution 3) Write out the complete solution : `particular` + `null`
(100,0)
Yep, keep going
how to find null soln
as the name says, it is the solution to the equation 7x + 13y = 0
give it a try.. it would feel awesome if you figure out a method to find the null solution on ur own..
(-13,-7)
Very close, but no. plug them in and see if they really produce 0
(-13,7)
Excellent! that is one null solution. Notice that any multiple of that also works, so all the null solutions are given by ` t(-13, 7)` where ` t ` belongs to the set of integers
100-13t>0 0+7t>0
Yes, you have skipped step3 but ok..
go ahead find the total count
what is step 3 ?
oh this one "3) Write out the complete solution : particular + null"
Yes, I was refering to that..
7 is answer
Yep! congratulations! Now you know how to solve any linear diophantine equation of form \(ax+by=c\)
cool!
so what was your method for finding null solution ?
ax + by = 0 (-b,a)
thats it!
that works always! so finding null solution is trivial as you can seethe only hard part is finding a particular solution
6y=200-4x 3y=100-2x \[y=\frac{ 100-2x }{ 3 }\] by hit and trial when x=2 \[y=\frac{ 100-4 }{ 3 }=32\] add successively 3 to the value of x x=2+3=5,y=30 x=5+3=8,y=28 x=11,y=26 .... x=47,y=2 x=50,y=0
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