Question

Three views of a stack of sugar cubes are shown below. What is the MINIMUM number of cubes that are required to produce this stack?

• A. 25
• B. 28
• C. 31
• D. 35

Hint:

Start with the $16$ cubes forced by the complete top view. Then overlap equal front-view and side-view peak requirements at common grid cells so that each extra vertical cube serves both silhouettes.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This question belongs to the topic of

3D Spatial Reasoning and Orthographic Projections.
We are given three orthographic views of a 3D stack of cubes: Top View, Front View, and Side View.
The goal is to determine the absolute minimum number of sugar cubes required to construct a stable 3D stack that satisfies all three views simultaneously.

Step 2: Key Formula or Approach:
We represent the stack as a 2D grid matching the Top View. Let the height of the cube stack at grid position \((i, j)\) be denoted by \(h_{i,j}\).

• The Top View shows a complete \(4 \times 4\) grid filled with squares, which means there must be at least one cube at every coordinate: \(h_{i,j} \ge 1\) for all \(1 \le i, j \le 4\).

• The Front View gives the maximum height of each column \(j\): \(\max_i h_{i,j}\). From left to right, these column maximums are: 4, 4, 3, 2.

• The Side View gives the maximum height of each row \(i\): \(\max_j h_{i,j}\). From back to front, these row maximums are: 4, 3, 2, 1.

• To minimize the total number of cubes, we must minimize the sum \(\sum_{i,j} h_{i,j}\) by overlapping the maximum height demands of rows and columns as much as possible.


Step 3: Detailed Explanation:

• Let us initialize all 16 cells in the \(4 \times 4\) grid with a height of 1 to satisfy the Top View requirement. This accounts for a base count of 16 cubes.

• Let us look at the row maximum requirements:
- Row 4 (front-most) has a maximum height of 1. Since all cells in this row are initialized to 1, this requirement is already satisfied.
- Row 3 has a maximum height of 2.
- Row 2 has a maximum height of 3.
- Row 1 (back-most) has a maximum height of 4.

• Let us look at the column maximum requirements:
- Column 1 has a maximum of 4.
- Column 2 has a maximum of 4.
- Column 3 has a maximum of 3.
- Column 4 has a maximum of 2.

• We can overlap these requirements strategically:
- Place a height of 4 at position \((1, 1)\). This satisfies Row 1 max (4) and Column 1 max (4) simultaneously.
- Place a height of 4 at position \((1, 2)\). This satisfies Row 1 max (4) and Column 2 max (4).
- Place a height of 3 at position \((2, 3)\). This satisfies Row 2 max (3) and Column 3 max (3) simultaneously.
- Place a height of 2 at position \((3, 4)\). This satisfies Row 3 max (2) and Column 4 max (2) simultaneously.

• All other cells can remain at their minimum height of 1. Let us write down the optimized height grid:
Row 1: [4, 4, 1, 1] (Max = 4)
Row 2: [1, 1, 3, 1] (Max = 3)
Row 3: [1, 1, 1, 2] (Max = 2)
Row 4: [1, 1, 1, 1] (Max = 1)

• Let us verify the column maximums for this grid:
Col 1: \(\max(4, 1, 1, 1) = 4\) (Satisfied)
Col 2: \(\max(4, 1, 1, 1) = 4\) (Satisfied)
Col 3: \(\max(1, 3, 1, 1) = 3\) (Satisfied)
Col 4: \(\max(1, 1, 2, 1) = 2\) (Satisfied)

• Let us sum the heights of all 16 cells in this optimized configuration:
\[ \text{Total Cubes} = (4 + 4 + 1 + 1) + (1 + 1 + 3 + 1) + (1 + 1 + 1 + 2) + (1 + 1 + 1 + 1) \]
\[ \text{Total Cubes} = 10 + 6 + 5 + 4 = 25 \]


Step 4: Final Answer:
The minimum number of cubes required to produce this stack is 25, which corresponds to option (A).

Answer 2

Concept:

• The top view requires a nonempty stack at every cell of a $4\times4$ footprint.
• The front and side silhouettes both require maximum heights $4,4,3,2$.
• Minimum construction is obtained by making the required row and column peaks coincide.

Step 1: Build the compulsory base layer.
All $16$ top-view cells are occupied, so begin with one cube in every cell: $16$ cubes.

Step 2: Align the two height-$4$ requirements.
Place height-$4$ stacks at two suitable intersections of height-$4$ rows and columns. Their extra cubes above the base contribute $2(4-1)=6$.

Step 3: Align the remaining peaks.
Place one height-$3$ stack at the intersection of the required height-$3$ row and column, adding $3-1=2$ cubes. Place one height-$2$ stack similarly, adding $2-1=1$ cube.

Step 4: Add the minimum contributions.
$16+6+2+1=25$. The aligned peaks reproduce $4,4,3,2$ from both viewing directions, so no additional cube is needed.

Final Answer: $25$, option A