A die has numbers from one to six on its face,  and the sum of the numbers on its opposite face is  7.  When six dice are placed as shown, with their faces touching each other having the same number  , what is the maximum sum of the numbers on the remaining faces of the dice, excluding the faces touching each other?

g
106

B.
102

C.
108

D.
104

Province
110

To solve this problem, let's break it down step by step:

### Step 1: Understanding the Dice Configuration
- The dice are placed such that the faces touching each other have the same number. 
- For any die, the sum of the numbers on opposite faces is 7.

### Step 2: Determine the Number on the Touching Faces
If two dice touch each other with the same number on their faces, then the opposite face of this touching face must have a number that sums to 7 with the touching number. For example:
- If the touching number is 1, the opposite number is 6.
- If the touching number is 2, the opposite number is 5.
- If the touching number is 3, the opposite number is 4.

### Step 3: Calculate the Maximum Sum of Remaining Faces
Since there are 6 dice, and each die has one face touching another die, we are left with 5 faces per die (excluding the touching face). We need to find the maximum sum of these 5 faces.

Given that opposite faces sum to 7, the highest possible number on any remaining face is 6, 5, or 4. 

### Step 4: Calculate the Total Sum
The sum of all numbers on a die is:
\[ 1 + 2 + 3 + 4 + 5 + 6 = 21 \]

If one face is touching and we want to maximize the sum of the remaining 5 faces, the worst-case scenario for the touching face would be the lowest number, i.e., 1, meaning the remaining 5 faces sum up to:
\[ 21 - 1 = 20 \]

Since the touching faces are the same across all 6 dice, the total maximum sum for the remaining faces is:
\[ 20 \times 6 = 120 \]

However, we need to subtract the sum of the 6 opposite faces from this sum, because the opposite of the touching face will also be counted.

For the worst-case scenario:
- If the touching face is 1, the opposite face is 6, so the sum of these opposite faces across all 6 dice is \( 6 \times 6 = 36 \).

Thus, the maximum sum of the remaining faces is:
\[ 120 - 36 = 84 \]

If the touching face is 6, the opposite face is 1, so:
\[ 120 - 6 = 114 \]

But for maximizing this, the configuration that maximizes the remaining sum would be when the touching face is 1 or any smaller number like 1 or 2.

But to keep things general, as it was established earlier, calculating by direct subtraction isn't right here:

So,
- \(5 \times 4 = 20\)
- \( 5 \times 5 = 25 \)
- \( 5 \times 6 = 30\)

Total Maximum sum is \(104\)

### Answer
D. 104

---

**Do you want more details or have any questions?**

### Related Questions:
1. How does the sum change if the dice are configured differently?
2. Can the total sum exceed 104 if the numbers are chosen differently?
3. What is the minimum sum of the numbers on the remaining faces?
4. How many different configurations of touching faces can lead to the same sum?
5. How does the symmetry of a die affect the outcome of such problems?

**Tip:** When solving problems involving dice or similar combinatorial objects, visualizing the configuration and understanding the relationships between the elements is crucial for maximizing or minimizing sums or other quantities.

Explore how to maximize the sum of remaining dice faces in a configuration where each die touches another with the same number. Understand the principles of dice configurations and the sum of opposite faces. Answer provided with step-by-step calculations and explanation. Suitable for Grades 7-9.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation