Least removed, most added

Here are the solid and the two views in the question. Solid A is built from 9 unit cubes standing on a table. More unit cubes are added to it, so that every cube still rests on the table or on another cube. What is the greatest number of cubes that can be added without changing the front view or the side view? Try it first. Pause if you like. Here is one possible arrangement of the nine cubes. We can solve this using horizontal layers. Lift the upper layers apart so we can see each layer. They will go back into the same places. At the bottom, the front view allows three cubes across. The side view allows three cubes from front to back. Fill the spaces in this bottom layer. The two views still fit. The bottom layer has three times three, or nine cubes. At the middle layer, the front view still allows three across. But the side view allows only one from front to back. This layer must stay along the front. The middle layer has three times one, or three cubes. At the top, the front view allows just one across. The side view also allows just one from front to back. The top layer has one times one, or one cube. Put the layers back. Both given views are unchanged, and no layer can hold another cube. Cubes when full equals nine plus three plus one equals thirteen. Cubes added equals thirteen minus nine equals four. Count the layers, then subtract the nine cubes we started with.

Solid B is built from 22 unit cubes. Some of the cubes are taken away, so that the solid that is left has the front view and the side view shown, and every cube left still rests on the table or on another cube. What is the least number of cubes that must be taken away? Try it first. Pause if you like. To remove the least, keep as many cubes as the two views allow. Separate the layers of Solid B so we can check what may stay. At the bottom, the front view allows three across. The side view allows three from front to back. All these cubes are already there. Keep three times three, or nine cubes, in the bottom layer. At the middle layer, the front view allows three across. The side view allows only two from front to back. Remove the back row from this layer. Keep the two rows at the front. Keep three times two, or six cubes, in the middle layer. At the top, the front view allows just one across. The side view allows just one from front to back. Keep the cube where those two views meet. Remove the other cubes in this layer. Keep one times one, or one cube, in the top layer. Put the remaining layers back. Both views fit, and we have kept every cube they allow. Cubes left equals nine plus six plus one equals sixteen. Cubes taken away equals twenty-two minus sixteen equals six.