A die is an unitary cube with numbers from $1$ to $6$ written on its faces, so that each number appears once and the sum of the numbers on any two opposite faces is $7$. We construct a large $3 \cdot 3 \cdot 3$ cube using$ 27$ dice. Find all possible values of the sum of numbers which can be seen on the faces of the large cube.
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Tags: geometry, combinatorics
rms
18.08.2024 19:51
parmenides51 wrote: A die is an unitary cube with numbers from $1$ to $6$ written on its faces, so that each number appears once and the sum of the numbers on any two opposite faces is $7$. We construct a large $3 \cdot 3 \cdot 3$ cube using$ 27$ dice. Find all possible values of the sum of numbers which can be seen on the faces of the large cube.
**Solution.** We call **type I dice** those that have only one visible face (those dice at the center of the faces of the large cube), **type II dice** those with two visible faces (those at the midpoints of the edges), and **type III dice** those with three visible faces (the corner dice). Any cube with an edge of 3 contains 6 type I dice, 12 type II dice, and 8 type III dice.
The minimum sum of the six faces is obtained when, on the type I dice, 1 is visible, on the type II dice, the visible faces are 1 and 2, and on the type III dice, the visible faces are 1, 2, and 3. The minimum sum is \( 6 \cdot 1 + 12 \cdot 3 + 8 \cdot 6 = 90 \).
The maximum is obtained when, on the type I dice, 6 is visible, on the type II dice, the visible faces are 5 and 6, and on the type III dice, the visible faces are 4, 5, and 6. The maximum sum is \( 6 \cdot 6 + 12 \cdot 11 + 8 \cdot 15 = 288 \).
We will show that the total sum can take any intermediate value between 90 and 288. For this, we start from the cube in which the minimum sum is realized and we will rotate the dice, step by step, so that we increase the sum by 1 each time.
By rotating a type I die, we can increase the total sum of the large cube's faces by one unit at each step, until the number 6 appears. We continue this process until all type I dice show the number 6.
Now we rotate a type II die from 1 + 2 to 5 + 6 and, at the same time, we rotate two of the type I dice from 1 to 4, respectively 4; thus, we have increased the total sum by 1. In the next steps, we rotate two more type I dice on which we acted, increasing the total sum by one unit each time; in the end, the number 6 will appear again. We continue this process so that all type II dice show faces 5 and 6.
Then, we rotate a type III die from 1 + 2 + 3 to 4 + 5 + 6, and at the same time, two of the type II dice from 1 to 2; thus, we have increased the total sum by 1. In the following steps, we rotate two more type I dice on which we acted, increasing the total sum by one unit each time; in the end, they will show non-6 faces. We continue this process until all type III dice show faces 4, 5, and 6.
We obtained a cube with 3 dice having the sum of the faces 288. In conclusion, the sum of the faces can take values from 90 to 288.
miyukina
19.08.2024 10:01
On this 3³ big die, there is one hidden small die, 6 face center dice, 12 edge center dice and 8 vertex dice { Max , min } of face centers = { 6, 1 } { Max , min } of edge centers = { 6 + 5, 1 + 2 } = { 11, 3 } { Max , min } of vertices = { 6 + 5 + 4, 1 + 2 + 3 } = { 15, 6 } { Max , min } sum of the 3³ die = { 1 × 0 + 6 × 6 + 12 × 11 + 8 × 15 , 1 × 0 + 6 × 1 + 12 × 3 + 8 × 6 } = { 0 + 36 + 132 + 120 , 0 + 6 + 36 + 48 } = { 288 , 90 } Answer : 90 ≤ sum of big die ≤ 288