Points in a Box

Consider the integral points ( x , y , z ) (x, y, z) on the plane 35 x + 55 y + 77 z = 1. 35 x + 55 y + 77 z = 1. How many are contained within a cube with side length 30 centered at ( 0 , 0 , 0 ) ? (0, 0, 0)?


The answer is 10.

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1 solution

Borut Levart
Sep 21, 2017

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To find the family of solutions, one can use the extended Euclidean algorithm . When solving by hand, I like to follow it in the following manner. First express the variable with the lowest factor:

x = ( 1 55 y 77 z ) / 35 x = (1 - 55 y - 77 z) / 35

We are looking for integral x x , so 35 should divide the numerator, or 35 1 + 15 y 7 z 35 | 1 + 15 y - 7 z , or 1 + 15 y 7 z = 35 k 1 + 15 y - 7 z = 35 k for any whole number k k . Take first step again, expressing the variable with the lowest factor:

z = ( 1 + 15 y 35 k ) / 7 z = (1 + 15 y - 35 k) / 7

Again, z z should be whole, so 7 1 + 15 y 35 k 7 | 1 + 15 y - 35 k , or 7 1 + y 7 | 1 + y , or y = 6 + 7 m y = 6 + 7 m for any whole number m m . So we have y y , plugging it above we get z z , and plugging both even higher above we get x x .

Part two: solving inequalities.

Can you show us (with more hints) on how this is done? I don't think your solution is particularly helpful.

Pi Han Goh - 3 years, 8 months ago

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Which part is troublesome, finding general solution or taking intersection with the box?

Borut Levart - 3 years, 8 months ago

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To be honest, I don't quite understand your question.

Are you saying that you want to construct a cube with one of its vertices at the origin (of a 3D Cartesian plane), and all its other vertices, that are also lattice points, are on the plane 35x+55y+77z=1 ?

Is my interpretation correct?

Pi Han Goh - 3 years, 8 months ago

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@Pi Han Goh The cube is centered at (0,0,0), has a side 30, which places two of its opposite corners at (-15,-15,-15) and (15, 15, 15). Then there are the points that lie on the specified plane ... We presume that the intersection of the plane points with the cube is not empty. But how many elements does that set have?

Borut Levart - 3 years, 8 months ago

One suggestion: I would note that the sides of the cube are included (in addition to the interior points) when computing the required integral triplets (i.e. x, y, z each lie in [-15, 15] ).

tom engelsman - 3 years, 8 months ago

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Thank you. I changed the question to: How many are contained within a cube with side 30 centered at { 0 , 0 , 0 } \{0, 0, 0\} ? I think it's now fairly reasonable to include any boundary points, do you agree?

Borut Levart - 3 years, 7 months ago

Bad solution

Laurence Kuo - 3 years, 4 months ago

My solution is a little more bashy. Because the gcd(77,55)=11, I found the values of x for which 1-35x divided 11. Beginning from -15, the first value of x is x=-5, where 1-35(-5)=176, which divides 11. We can reason that -5+11 will work (which indeed it does.) At this point, we have the 2 integer values of x for which we can find solutions (hopefully) within the boundaries. We have 35(-5) + 55y + 77z = 1 and 35(6) + 55y+77z = 1. Which can be simplified to 16=5x+7y; -19=5x+7y. and we find 10 solutions within the boundaries.

Aaron Lin - 1 year, 11 months ago

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