I agree with you but there is no such information given .
We can assume the larger brick should be made using smaller bricks only without concerning its size .
but we can see the answer varies with the size of larger brick ,see
we can understand the two sides of the large brick is a multiple of 10 .(since smaller brick has sides of 10*20)
remaining side of the larger brick is at least a multiple of 5 .
since there is no information given about the larger brick we can select a smaller brick which has a side of 5 i.e brick
It's all to do with the orientation of the small bricks -which way they are pointing. If we imagine fixed x, y and z axes and a small brick is placed so that its plane faces are parallel to the y-z, z-x and x-y planes then there are six possible orientations of the brick. A large brick is then made up of an unknown number of small bricks placed in various orientations, but (presumably) without leaving any gaps. The problem is then to prove that we can dismantle the large brick and re-arrange the small ones to re-create the large one, in such a way that all the small bricks are 'pointing the same way'. This is what you meant by
As you said, this amounts to proving that the sides of the large brick are multiples of 5, 10 and 20. But I can't see how to do that at present.all edges of equal lengths being parallel.