Jonas builds a large $n \times n \times n$ cube using $1 \times 1 \times 1$ cubes each having the net shown. What is the smallest value of $n$ for which the sum of the exterior faces of the $n \times n \times n$ cube can be greater than 1500 ?
(A) 9
(B) 11
(C) 12
(D) 13
(E) 16

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Let the sum of the numbers on the exterior faces of the $n \times n \times n$ cube be $S$.

To determine the smallest value of $n$ for which $S>1500$, we choose to position the $1 \times 1 \times 1$ cubes within the large cube in such a way that $S$ is as large as possible.

The $1 \times 1 \times 1$ cubes which contribute to the numbers on the exterior faces of the large cube can be classified as one of three types.

We call these three types: corner, edge and inside.

In the portion of the large $n \times n \times n$ cube shown in the diagram below, each of these three types is shown.

(i) A corner cube is shown in Figure 1. These are cubes that appear in one of the "corners" of the large cube and so there are 8 such corner cubes.

(ii) An edge cube is shown in Figure 2. These are cubes that appear along the edges but not in the corners of the large cube.

A cube has 12 edges and each edge of the large cube contains $n-2$ edge cubes, and so there are $12 \times(n-2)$ such cubes.

(iii) An inside cube is shown in Figure 3. These are the remaining cubes that contribute to the numbers on the exterior faces of the large cube.

A cube has 6 faces and each face of the large cube contains $(n-2) \times(n-2)$ inside cubes, and so there are $6 \times(n-2)^{2}$ such cubes.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-061.jpg?height=385&width=1434&top_left_y=146&top_left_x=432)

Each corner cube has 3 faces which contribute to $S$.

For $S$ to be as large as possible, the $1 \times 1 \times 1$ cube must be positioned in such a way that the sum of the 3 external faces is as large as possible.

We may determine from the given net that the three faces meeting at a vertex of the $1 \times 1 \times 1$ cube may contain the numbers: $-1,0,1$ or $-1,2,0$ or $-2,2,0$ or $-2,1,0$.

The sums of these three faces are $0,1,0$, and -1 respectively.

To make $S$ as large as possible, we choose to place each corner cube so that the numbers appearing on the exterior faces of the large cube are $-1,2$ and 0 , and thus have the largest possible sum of 1 .

Therefore, the 8 corner cubes contribute $8 \times 1=8$ to $S$.

Each edge cube has 2 faces which contribute to $S$. For $S$ to be as large as possible, the $1 \times 1 \times 1$ cube must be positioned in such a way that the sum of the 2 external faces is as large as possible.

We may determine from the given net that two faces which share an edge of the $1 \times 1 \times 1$ cube contain the numbers: $-1,0$ or $-1,2$ or $-1,1$ or 2,0 or 1,0 or $-2,0$ or $-2,1$ or $-2,2$.

The sums of these two adjacent faces are $-1,1,0,2,1,-2,-1$, and 0 respectively.

To make $S$ as large as possible, we choose to place each edge cube so that the numbers appearing on the exterior faces of the large cube are 2 and 0 , and thus have the largest possible sum of 2 . Therefore, the $12 \times(n-2)$ edge cubes contribute $2 \times 12 \times(n-2)$ or $24 \times(n-2)$ to $S$.

Finally, each inside cube has 1 face which contributes to $S$. For $S$ to be as large as possible, we position each of these cubes so that 2 will appear on this face (since 2 is the largest number in the given net).

Thus, the $6 \times(n-2)^{2}$ inside cubes contribute $2 \times 6 \times(n-2)^{2}$ or $12 \times(n-2)^{2}$ to $S$.

In total, we get $S=8+24 \times(n-2)+12 \times(n-2)^{2}$.

We want the smallest value of $n$ for which $S>1500$ or $8+24 \times(n-2)+12 \times(n-2)^{2}>1500$. Using trial and error with the given answers, we get the following:

- When $n=9, S=8+24 \times(n-2)+12 \times(n-2)^{2}=8+24 \times 7+12 \times 7^{2}=764$ which is less than 1500 .
- When $n=11, S=8+24 \times 9+12 \times 9^{2}=1196$ which is less than 1500 .
- When $n=12, S=8+24 \times 10+12 \times 10^{2}=1448$ which is less than 1500 .
- When $n=13$, $S=8+24 \times 11+12 \times 11^{2}=1724$ which is greater than 1500 .

The smallest value of $n$ for which the sum of the exterior faces of the $n \times n \times n$ cube is greater than 1500 is 13 .