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First create your equation of your line

y = mx + c
slope of PQ is -3/5 and intercept is 30

Hence y = -(3/5)x + 30
5y + 3x = 150
Find the calue of y and x such that 5y + 3x adds up to 150
So if x is 0 then 5(30) = 150
x is 5 then 5(27) = 135

so divide 50/5 and add one for zero = 10 + 1 = 11
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Alternative approach:
Number of integer points can be calculated as GCD(30,50)+1=11.
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greenoak
Alternative approach:
Number of integer points can be calculated as GCD(30,50)+1=11.
this seems to be a promising solution.
pl confirm if it can be applied to other similar problems?
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Quote:
this seems to be a promising solution.
pl confirm if it can be applied to other similar problems?

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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