a. 

There are many possible answers; two possibilities are 2 by 2 by 8 and 2 by 4 by 4. These are found by doubling any one of the three dimensions of the original.

b. 

Of the many possible answers, two possibilities are 2 by 4 by 8 and 4 by 4 by 4. These are found by doubling any two of the three dimensions (or multiplying one dimension by a factor of 4).

c. 

Of the many possible answers, a 4-by-4-by-8 box is similar to the original. This is formed by doubling all three of the original's dimensions.

a. 

Only four of Package 1 will fit in Box B.
Sixteen of Package 2 will fit in Box B (vertically), filling the box.
Twelve of Package 3 will fit in Box B, filling the box.
Four of Package 4 will fit in Box B, filling the box.
Package 5 will not fit in Box B at all. Its largest dimension is larger than any of the dimensions of the box!

b. 

Answers will vary, but one strategy is to align the packages along matching dimensions. Recognizing that a package is 1 by 1 by 3 helps to fit it into a 4-by-4-by-3 box.

a. 

One approach is to think about the dimensions of the new box with respect to the dimensions of Packages 1-5. For example, one dimension of the new box needs to be divisible by 5 (because of Package 5). We need all three dimensions to be divisible by 2 (because of Package 1). We need one dimension divisible by 3 (because of Package 2 and Package 4). Based on these observations, the dimensions of the new box are 2 by (2 • 5) by (2 • 3), which is 2 by 10 by 6. Using this same reasoning, you could also decide on a 2-by-2-by-30 box (which also works for all five packages). These are the smallest (in total volume) that will work.

b. 

We know that a 2-by-10-by-6 box works. If we double any of the sides, for example, then the dimensions are all still divisible by the necessary lengths, so it will work to ship all of the packages. By this reasoning, if the sides of a box are "divisible" by 2, 10, and 6, or "divisible" by 2, 2, and 30 from our second example, then it will work to ship all of the packages.

c. 

The dimensions of the larger box must be divisible by the dimensions of the smaller packages. Given a small package with dimensions l, w, and h, we must have one dimension of the box divisible by l, a different dimension divisible by w, and the third dimension divisible by h. For example, for Package 1 to completely fill the box, we see that each dimension must be divisible by 2 (i.e., it must be even).

d. 

Each dimension of the box has to be a common multiple of unique dimension of each of the packages. So for any package with dimensions l, w, and h, we must have one dimension of the box divisible by l, a different dimension divisible by w, and the third dimension divisible by h.