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Line segment wih points as integers
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chidcguy
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Please see attachment. I solved it but it took around 4 min and I would like to know if there is an easy method. Thanks
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Ian Stewart
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The y-intercept of the line is 30, and the slope is -30/50. So the equation of the line is:
y = -(3/5)x + 30
There are 51 integers we can plug in for x (anything from 0 to 50), but we'll only get an integer value of y if x is a multiple of 5 (otherwise the '5' in the denominator of 3/5 won't cancel). So the values of x that will give us integer co-ordinates are 0, 5, 10, 15 ..., 45, 50 = 0*5, 1*5, 2*5, ..., 9*5, 10*5, or eleven values in total.
y = -(3/5)x + 30
There are 51 integers we can plug in for x (anything from 0 to 50), but we'll only get an integer value of y if x is a multiple of 5 (otherwise the '5' in the denominator of 3/5 won't cancel). So the values of x that will give us integer co-ordinates are 0, 5, 10, 15 ..., 45, 50 = 0*5, 1*5, 2*5, ..., 9*5, 10*5, or eleven values in total.
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chidcguy
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Ian, Thats what I did.
Solve for slope m and formed the line 3x+5y=150 and then realized that the only values 5y will end in are the values that will have a 0 or 5 at the end and enumerated all of them. Took some time to figure out the 0/5 part for 5y.
Solve for slope m and formed the line 3x+5y=150 and then realized that the only values 5y will end in are the values that will have a 0 or 5 at the end and enumerated all of them. Took some time to figure out the 0/5 part for 5y.
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