When any odd number greater than unity multiplied by even times by itself than dividing this product by 8, we get the remainder as :

Options

A. a. 1
B. b. 7
C. c. not unique
D. d. none of these

Answer

To solve the problem, we need to analyze the situation where an odd number greater than 1 is multiplied by itself an even number of times, and then the product is divided by 8 to find the remainder.

### Step-by-step Solution:

1. **Identify an Odd Number**:
Let's take an odd number greater than 1. For example, we can choose \( n = 3 \).

2. **Multiply the Odd Number by Itself an Even Number of Times**:
Since we need to multiply it an even number of times, we can take \( n^2 \) (which is multiplying it by itself once).
\[ n^2 = 3^2 = 9 \]

3. **Divide the Product by 8**:
Now, we need to divide the result by 8 and find the remainder.
\[ 9 \div 8 = 1 \quad \text{(remainder 1)} \]

4. **Try Another Odd Number**:
Let's try another odd number, say \( n = 5 \).
\[ n^2 = 5^2 = 25 \]
Now divide by 8:
\[ 25 \div 8 = 3 \quad \text{(remainder 1)} \]

5. **Generalize the Result**:
We can generalize this for any odd number \( n \). The square of any odd number can be expressed as:
\[ n = 2k + 1 \quad \text{(where \( k \) is an integer)} \]
Then,
\[ n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 4k(k + 1) + 1 \]
Since \( k(k + 1) \) is always even, \( 4k(k + 1) \) is divisible by 8. Thus, when we divide \( n^2 \) by 8, the remainder is:
\[ 1 \]

### Conclusion:
For any odd number greater than 1, when multiplied by itself an even number of times and divided by 8, the remainder is always **1**.