Lattice points are points whose coordinates are integers. The unit cube is a cube whose sides have a length of one unit. How many lattice points are on, or enclosed within, a unit cube whose vertices are lattice points?

A. 6 B. 8 C. 9 D. 16 E. 32

(B) The only difficulty in solving this problem is in the visualization of the unit cube. However, this difficulty can easily be resolved once you realize that we are working with a unit cube. A unit cube has sides whose lengths are equal to 1. So, for example, if the cube had one vertex at the origin, (0, 0, 0), and was positioned in the first quadrant, another point would be (1, 0, 0).

Essentially, what you are prompted to realize is that the number of lattice points on or within the unit cube is exactly equal to the number of vertices on the unit cube, and any cube has a total of 8 vertices. So the total number of lattice points within the cube is equal to 8, or answer choice (B)

I really don't get how this question is solved. Please, explain how the cube is aligned?

Post subject: Re: math: x,y,z - space, lattice points.

Posted: Wed May 12, 2010 1:02 pm

Joined: Fri Apr 09, 2010 2:11 pm Posts: 453

This question is solely based on 3D-imagination. However there is a path we can go to make it easier.

First, let us start with a 2D. Lattice points on x,y-plane (2D) are points whose coordinates (x,y), where x, y are integers. Here they are:

Analogue of the unit cube in 2D is a unit square. Let us choose one arbitrary lattice point as one of the vertices of the square. Let it be point (3, 2), for example:

Another vertex of the square is 1 unit away from this one because the square has 1-unit sides (as in 3D, cube has 1-unit edges). So the second vertex must lie on the 1-unit circle whose center is point (3, 2):

You can see that this circle crosses 4 other lattice points: (3, 1), (3, 3), (1, 2), (4, 2). So these are the only options for another vertex. In either case 4 possible options for a square, once we choose second vertex, has lattice points as vertices and none inside or on the sides:

So everything is clear with 2D so far. Let us move on to 3D. First, imagine all lattice points (x, y, z), where x, y, z are integers:

Then let us chose one arbitrary point. For example, let it be (0, 0, 0).

Another vertex is one unit away from (0, 0, 0) so it lies on the 1-unit sphere.

In any case we have 1-unit cube whose vertices are lattice points and none inside or on the edges.

So the answer is 8 lattice points (vertices of a cube), choice (B).

The key to solve this question is to understand that nearby vertices have 1 unit change in only one of the coordinates and no other option possible (remember 1-unit circle and 1-unit sphere we drew).

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