Integral Points
1. The lines y=3x and x+y=40 and the x axis bound a triangular area . Find the total number of points on or inside this triangle with integral coordinates?
Can anyone, give the complete solution of this problem and point out a general way(if any), for solving these type of questions.
But 230 is not in the option,(its the first question of the first Math FLT), so I think its not the correct solution.
Moreover, I would like to correct you at one part of your solution i.e,
"Along X axis i.e when x =0 we have 40 points
(0,0), (0,1), . . . . .(0,40)"
It is 41 points along the x-axis inclusive of both start and end ponts.
A bit busy now in office and ‘ll able to write the solution after reaching home may be after 10pm
The answer is 655
n/a
655 is what is given in the key.
answer i am getting is 641.
Anyone else getting 641?
First lets try to find out all the three vertices of the triangle
1. (0 ,0) let it be O
2. Intersection of x axis and x + y = 40 is (0, 40) let this point be A.
3.Intersection of y=3x axis and x + y = 40 is (10, 30) let this point be B.
We have to find how many integral points are with in these three lines.
Lets start from x axis
Along X axis i.e when x =0 we have 40 points
(0,0), (0,1), . . . . .(0,40)
when x =1 along y =3x we have y=3 and that’s the start point and end point is on line x + y =40 then when y =39. So the total number of points from 3 to 39 is 37.
When x =2 , y= 6 start point and y = 38 is end point so total number of points 33.
When x =3 , y= 9 start point and y = 37 is end point so total number of points 29.
Like so
When x= 10, y =30 start point and y=30 is end point so total number of points 1.
So sum of all the possible points= 40 + 37 + 33+ 29 + …1
= 40 + 10/2(2 x 1 + 9 x 4 )
= 40 + 38 x 5
= 230
Hopefully I have not done any calculation mistake.