Question

use the following magic multiplication square to answer the following parts (a)-(c).

\

$$\begin{array}{|c|c|c|} \\hline 8 & 256 & 2 \\\\ \\hline 4 & 16 & 64 \\\\ \\hline 128 & 1 & 32 \\\\ \\hline \\end{array}$$

select the correct choice below and, if necessary, fill in any answer box that completes your choice.

a. after rewriting the numbers using the same base, 2, the sum of the bases for each row, column, and diagonal is .
b. after rewriting the numbers using the same base, 2. the numbers can be rearranged to form the same sum for each row, column, and diagonal. that sum is .
c. after rewriting the numbers using the same base, 2, the sum of the exponents for each row, column, and diagonal is 12.

(c) create a magic multiplication table with the numbers 1, 3, 9, 27, 81, 243, 729, 2187, and 6561. three numbers have been entered as shown on the right. fill in the rest of the magic multiplication square.

\

$$\begin{array}{|c|c|c|} \\hline 243 & & \\\\ \\hline & & 9 \\\\ \\hline & 6561 & \\\\ \\hline \\end{array}$$

Explanation:

Analyze the properties of the base-2 magic square

The first part of the image shows a magic multiplication square:
Row 1: \(8, 256, 2\)
Row 2: \(4, 16, 64\)
Row 3: \(128, 1, 32\)

Rewriting each number as a power of base \(2\):
Row 1: \(2^3, 2^8, 2^1\)
Row 2: \(2^2, 2^4, 2^6\)
Row 3: \(2^7, 2^0, 2^5\)

The exponents form a standard addition magic square:
Row 1: \(3, 8, 1\) (sum = \(12\))
Row 2: \(2, 4, 6\) (sum = \(12\))
Row 3: \(7, 0, 5\) (sum = \(12\))

Thus, option C is correct: "After rewriting the numbers using the same base, 2, the sum of the exponents for each row, column, and diagonal is 12."

Determine the target sum of exponents for base-3 square

Part (c) asks to create a magic multiplication table with powers of 3:
\(1, 3, 9, 27, 81, 243, 729, 2187, 6561\)
These are \(3^0, 3^1, 3^2, 3^3, 3^4, 3^5, 3^6, 3^7, 3^8\).

The exponents are \(0, 1, 2, 3, 4, 5, 6, 7, 8\).
The sum of these exponents is:

$$0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36$$

Since there are 3 rows, the magic sum of exponents for each row, column, and diagonal must be:

$$\frac{36}{3} = 12$$

The center cell of the exponent magic square must be the average of the exponents:

$$\frac{12}{3} = 4$$

Thus, the center cell of the multiplication square is \(3^4 = 81\).

Map the given values to exponents

Let the exponent magic square be:

$$
LATEXBLOCK0
$$

We are given the following values in the multiplication square:

• Row 1, Column 1: \(243 = 3^5 \implies a = 5\)
• Row 2, Column 3: \(9 = 3^2 \implies f = 2\)
• Row 3, Column 2: \(6561 = 3^8 \implies h = 8\)

We also know the center exponent \(e = 4\) (since \(3^4 = 81\)).

Solve the exponent magic square

Using the magic sum of \(12\):

• For Row 2: \(d + e + f = 12 \implies d + 4 + 2 = 12 \implies d = 6\)
• For Column 2: \(b + e + h = 12 \implies b + 4 + 8 = 12 \implies b = 0\)
• For Row 1: \(a + b + c = 12 \implies 5 + 0 + c = 12 \implies c = 7\)
• For Column 1: \(a + d + g = 12 \implies 5 + 6 + g = 12 \implies g = 1\)
• For Column 3: \(c + f + i = 12 \implies 7 + 2 + i = 12 \implies i = 3\)

Let's verify the diagonals:

• Diagonal 1: \(a + e + i = 5 + 4 + 3 = 12\) (Correct)
• Diagonal 2: \(g + e + c = 1 + 4 + 7 = 12\) (Correct)

The exponent magic square is:

$$
LATEXBLOCK1
$$

Convert exponents back to base-3 values

Now we convert each exponent back to its corresponding power of 3:

• Row 1:
• \(3^5 = 243\)
• \(3^0 = 1\)
• \(3^7 = 2187\)
• Row 2:
• \(3^6 = 729\)
• \(3^4 = 81\)
• \(3^2 = 9\)
• Row 3:
• \(3^1 = 3\)
• \(3^8 = 6561\)
• \(3^3 = 27\)

Answer:

Part (b)

• A. After rewriting the numbers using the same base, 2, the sum of the bases for each row, column, and diagonal is [ ]
• B. After rewriting the numbers using the same base, 2. The numbers can be rearranged to form the same sum for each row, column, and diagonal. That sum is [ ]
• C. After rewriting the numbers using the same base, 2, the sum of the exponents for each row, column, and diagonal is 12 (Correct answer)

Part (c)

The completed magic multiplication square is:

243	1	2187
3	6561	27