If m is even, then (8^m - 1) is divisible by which of the following?

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Given m is even, let m = 2k for some integer k.

We want to check divisibility of (8^m - 1) = (8^(2k) - 1).

Note that 8 = 2^3, so 8^(2k) = (2^3)^(2k) = 2^(6k).

We want to find a number that divides (8^(2k) - 1).

Let's analyze option 1: 63.  
63 = 7 * 9.

Check divisibility by 7:  
Since 8 ≡ 1 (mod 7) because 8 - 7 = 1,  
8^(2k) ≡ 1^(2k) ≡ 1 (mod 7).  
So, 8^(2k) - 1 ≡ 0 (mod 7).

Check divisibility by 9:  
Since 8 ≡ -1 (mod 9) because 8 - 9 = -1,  
8^(2k) ≡ (-1)^(2k) ≡ 1 (mod 9) because 2k is even.  
So, 8^(2k) - 1 ≡ 0 (mod 9).

Since (8^(2k) - 1) is divisible by both 7 and 9, it is divisible by their product 63.

Check other options:  
- 42 = 2 * 3 * 7. Since 8^(2k) - 1 is odd, it cannot be divisible by 2.  
- 65 = 5 * 13. 8^(2k) mod 5 and mod 13 do not consistently yield 1.  
- 8 is a power of 2, but 8^(2k) - 1 is always odd, so not divisible by 8.

Therefore, the correct answer is 63.

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