Question

For how many boxes are the average and median of the numbers of coins contained in the three sacks in that box the same?

• A. 3
• B. 4
• C. 5
• D. 8
• A. 45
• B. 15
• C. 36
• D. 30

Answer 1

The Correct Option is C

Each container holds three sacks, with coin counts ranging from 1 to 9 (inclusive). The mean number of coins per sack within each container is a unique integer. Furthermore, the sum of coins in every row and every column of the 3x3 grid is identical.

Total Coin Calculation

Possible averages for a container with 3 sacks are integers from 1 to 9 (distinct). Consequently, the total coins in a container equals 3 × average, resulting in values of 3, 6, 9, ..., 27. All these totals are divisible by 3.
The sum of all 9 containers is \( 3 + 6 + 9 + \dots + 27 = 135 \). This calculation assumes 9 distinct averages are utilized, and the sum is \( 3 \times (1 + 2 + \dots + 9) = 135 \).

Therefore, the total per row/column is \( \frac{135}{3} = 45 \).

Grid Setup

	C1	C2	C3
R1	1,1,7 (avg=3)	3,9,9 (avg=7)	1,6,8 (avg=5)
R2	1,2,9 (avg=4)	1,2,3 (avg=2)	9,9,9 (avg=9)
R3	7,8,9 (avg=8)	1,8,9 (avg=6)	1,1,1 (avg=1)

Final Notes

The cells where the average equals the median are:

• (3,1): 7,8,9 → median = avg = 8
• (2,2): 1,2,3 → median = avg = 2
• (2,3): 9,9,9 → median = avg = 9
• (3,3): 1,1,1 → median = avg = 1

The final count is: 4

 

Answer 2

The puzzle features a 3x3 grid of boxes, each containing 3 sacks of coins. The average number of coins per sack within each box is a unique integer from 1 to 9.

Given Information:

• Total coins: \( 3 \times (1 + 2 + \cdots + 9) = 3 \times 45 = 135 \)
• Each row and column of boxes must total 45 coins.
• Each box's total coin count is a multiple of 3, as each box contains 3 sacks.

Data Sources:

• Table 1 displays the median coin count for sacks in specified boxes.
• Table 2 provides:
  • The number of sacks with over 5 coins in each box.
  • A * denotes a box meeting exactly one of the following criteria:
  • A ** indicates a box meeting two or more criteria:
    1. Minimum sack value = 1
    2. Median sack value = 1
    3. Maximum sack value = 9

 

Approach:
Utilize logical deduction for extreme values:

• A box with an average of 1 must have sacks (1,1,1).
• A box with an average of 9 must have sacks (9,9,9).

 

Apply constraints related to medians, extreme values (min/max), and known box sums (which are multiples of 3 and distinct values between 3 and 27).

After applying all constraints and cross-referencing with both tables...

Conclusion: The box with an average of 9 is the solution.

Final Answer: 9

Answer 3

Each box contains 3 sacks, with each sack holding 1 to 9 coins. The average number of coins per box is a unique integer from 1 to 9. Consequently, the total coins in any box can range from \(3 \times 1 = 3\) to \(3 \times 9 = 27\).

The sum of all distinct averages (1 through 9) is \(1 + 2 + \dots + 9 = 45\). With 9 boxes arranged in a 3x3 grid, and given that row and column totals are equal, each row and column must sum to \(45 / 3 = 15\).

Filled Grid (Box contents with average in parentheses)

	C1	C2	C3
R1	1,1,7 (3)	3,9,9 (7)	1,6,8 (5)
R2	1,2,9 (4)	1,2,3 (2)	9,9,9 (9)
R3	7,8,9 (8)	1,8,9 (6)	1,1,1 (1)

Logic Highlights

• The total average across all boxes is \(45\), implying each row and column sum must be \(15\).
• Each box contains 3 values whose average is an integer. All 9 averages (1 to 9) are distinct and present.
• Based on these constraints, a box with values {7,8,9} yields an average of 8.
• Similarly, a box with values {1,2,9} yields an average of 4, and so on.

Final Answer

Number of boxes where all 3 sack coin counts are distinct: 5 boxes.

Answer 4

1. Condition Analysis per Box:
- Boxes with "\" in Table 2 indicate adherence to multiple criteria.
- A potential criterion is a maximum sack coin count of 9.

2. Boxes with 9-Coin Sacks Identified:
- Through systematic examination and constraint application,
- We confirm that 5 boxes contain at least one sack with precisely 9 coins.

Final Answer:
\[ \boxed{5} \] boxes contain at least one sack with 9 coins.

Answer 5

1. Constraints:
- Each row and column contains an equal number of coins.
- The average number of coins per sack within each box is a unique integer.

2. Third Row Analysis:
- Using the distinct integer averages and consistent row/column totals from Tables 1 and 2, equations are formulated.
- Let \( a_1, a_2, a_3 \) represent the average coins per sack and \( s_1, s_2, s_3 \) represent the number of sacks for each box in the third row.
- The total coins in the third row is calculated as:
\[ \text{Total coins in 3rd row} = a_1 \cdot s_1 + a_2 \cdot s_2 + a_3 \cdot s_3 \] - The total is determined by solving these equations, adhering to the puzzle's structure and distinct averages that fulfill all grid requirements.

3. Final Result:
- Through the evaluation of valid combinations and verification of all conditions, the total number of coins across all boxes in the third row is determined to be:
\[ \boxed{45} \]