The highest power of 17 which can divide exactly the following expression:
`(18^2 - 1) (18^4- 1)(18^8 - 1)(18^10 - 1) xx ……. (18^(16) - 1)(18^(16) - 1)(18^(18) - 1)` is :

Options

A. 1
B. 17
C. 9
D. can't be determined

Answer

To find the highest power of 17 that can divide the expression \((18^2 - 1)(18^4 - 1)(18^8 - 1)(18^{10} - 1) \cdots (18^{16} - 1)(18^{18} - 1)\), we can follow these steps:

### Step 1: Identify the expression
The expression consists of several factors of the form \(18^n - 1\) where \(n\) takes the values: 2, 4, 8, 10, 12, 14, 16, and 18.

### Step 2: Factor each term
Each term \(18^n - 1\) can be factored using the difference of squares or the difference of powers:
- For even \(n\), \(18^n - 1 = (18^{n/2} - 1)(18^{n/2} + 1)\).
- For \(n = 2k\), \(18^n - 1\) is divisible by \(18 - 1 = 17\) and \(18 + 1 = 19\).

### Step 3: Count the factors of 17
We need to determine how many of the terms \(18^n - 1\) are divisible by 17. Since \(18 \equiv 1 \mod 17\), we can conclude that:
- \(18^n - 1 \equiv 0 \mod 17\) for all even \(n\).

### Step 4: List the powers of n
The values of \(n\) we have are: 2, 4, 8, 10, 12, 14, 16, and 18. This gives us a total of 9 terms.

### Step 5: Conclusion
Since each of the 9 terms contributes at least one factor of 17, the highest power of 17 that divides the entire expression is \(17^9\).

### Final Answer
The highest power of 17 which can divide the expression exactly is \(9\).

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