Answer the following: a. If 6 x A = 2A, find the value of A. b. If C x C = 3C, find the value of C. c. If A7 x A = B4, find the possible values of A and B. d. If AB x B = 261, find... Answer the following: a. If 6 x A = 2A, find the value of A. b. If C x C = 3C, find the value of C. c. If A7 x A = B4, find the possible values of A and B. d. If AB x B = 261, find the value of A and B. e. Create a 3 x 3 magic square of magic constant 24. Create a 4 x 4 magic square using the numbers given below: 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47 Read more

Steps to Solve

1. Solve for A in 6 x A = 2A

To solve the equation $6 \times A = 2A$, we can rearrange the equation: $6A = 2A$ Subtract $2A$ from both sides: $6A - 2A = 0$ $4A = 0$ Divide both sides by 4: $A = 0$

2. Solve for C in C x C = 3C

To solve the equation $C \times C = 3C$, we can rewrite it as: $C^2 = 3C$ $C^2 - 3C = 0$ Factor out C: $C(C - 3) = 0$ This gives two possible solutions for C: $C = 0$ or $C - 3 = 0$, which means $C = 3$.

3. Solve for A and B in A7 x A = B4

We are looking for a value of A such that when multiplied by A7 (which is $10A + 7$), we get a number in the form B4 (which is $10B + 4$). $(10A + 7) \times A = 10B + 4$ Let's test values of $A$ from 1 to 9. If $A = 1$, $(10(1) + 7) \times 1 = 17 \times 1 = 17$. This is not in the form B4. If $A = 2$, $(10(2) + 7) \times 2 = 27 \times 2 = 54$. So, $B = 5$ and the number is 54. Thus, $A=2$, $B=5$ is a possible solution If $A = 3$, $(10(3) + 7) \times 3 = 37 \times 3 = 111$. This is not in the form B4. If $A = 4$, $(10(4) + 7) \times 4 = 47 \times 4 = 188$. This is not in the form B4. If $A = 5$, $(10(5) + 7) \times 5 = 57 \times 5 = 285$. This is not in the form B4. If $A = 6$, $(10(6) + 7) \times 6 = 67 \times 6 = 402$. This is not in the form B4. If $A = 7$, $(10(7) + 7) \times 7 = 77 \times 7 = 539$. This is not in the form B4. If $A = 8$, $(10(8) + 7) \times 8 = 87 \times 8 = 696$. This is not in the form B4. If $A = 9$, $(10(9) + 7) \times 9 = 97 \times 9 = 873$. This is not in the form B4. Therefore, the only solution is $A = 2$ and $B = 5$.

4. Solve for A and B in AB x B = 261

Here, AB represents the number $10A + B$. So, the equation becomes: $(10A + B) \times B = 261$ We need to find integers A and B (where A and B are digits from 0 to 9) that satisfy this equation. We can rewrite this as: $10AB + B^2 = 261$ We can rearrange this as: $B(10A + B) = 261$ Since B is a factor of 261, we can find the factors of 261. The factors of 261 are 1, 3, 9, 29, 87, and 261. Since B is a single digit, the possible values for B are 1, 3, and 9.

If $B = 1$, then $10A + 1 = 261$, so $10A = 260$, and $A = 26$. This is not possible since A must be a single digit. If $B = 3$, then $10A + 3 = 261/3 = 87$, so $10A = 84$, and $A = 8.4$. This is not possible since A must be an integer. If $B = 9$, then $10A + 9 = 261/9 = 29$, so $10A = 20$, and $A = 2$. So, $A = 2$ and $B = 9$ is a solution.

5. Create a 3 x 3 magic square of magic constant 24

A magic square has rows, columns, and diagonals that sum to the same "magic constant". With a 3x3 grid, it will contain 9 numbers. Since the magic constant is 24, we need nine numbers such that the rows, columns, and diagonals sum up to 24. One possible solution is:

6. Create a 4 x 4 magic square using the numbers 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47

The numbers given are an arithmetic sequence with a common difference of 3. First, determine the magic constant. The sum of all the numbers is: $2 + 5 + 8 + 11 + 14 + 17 + 20 + 23 + 26 + 29 + 32 + 35 + 38 + 41 + 44 + 47 = 408$ Since there are 4 rows, the magic constant is $408 / 4 = 102$.

Here's one possible magic square: