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PQ is a straight line passing through (50,0) and (0,30). Slope of line = (30-0)/(0-50) = -3/5 Y intercept = 30 Equation of line = y = mx + c, y = -3/5x + 30 => 3x+5y = 150
Now you need only value for x and y those are integer values and those that will satisfy this equation. Moreover you cannot have -ve values and for x max value is 50 and and for y it is 30.
For x = 0, 5, 10, 15, ..., 50 you can have integer Y ranging from 30 to 0.
this seems to be a promising solution. pl confirm if it can be applied to other similar problems?
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It seems that yes, it can be. Here is my reasoning:
Let x/A+y/B=1 be the equation of the line (A, B are non-zero integers). Let’s take an integer n: 1 y=B(n-1)/n.
So, we need to find n not greater than A such that both A/n and B(n-1)/n are integers. B(n-1)/n is an integer B – B/n is integer B/n is integer.
It means we need to find the greatest possible n such that both A/n and B/n are integers (for such an n, A/n will correspond to the smallest coordinate x such that the point (x,y) on the line will be integer. x-coordinates of other integer points will be 2/n*A, 3/n*A, ..., (n-1)/n*A, A). And, by definition, the greatest possible n such that A/n, B/n are integers is GCD(A,B).
We need to add 1 because the case (0,B) also gives us an integer, but we did not count that point.
Therefore, in general case, the formula for the number of integer points on the line segment is N = GCD(A,B)+1.
*** I'd be grateful for other people's comments on the reliability of this reasoning.
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