If n is a positive integer and `(3sqrt(3)+5)^(2n+1)=l+f` where l is an integer annd `0 lt f lt 1`, then

Options

A. `alpha ` is an even integer
B. `(alpha + beta)^(2)` is divisible by `2^(2n+1)`
C. the integer just below `(3sqrt(3) + 5)^(2n+1)` divisible by 3
D. `alpha ` is divisible by 10

Explanation

To solve the problem, we start with the expression given:

\[ (3\sqrt{3} + 5)^{2n + 1} = l + f \]

where \( l \) is an integer and \( 0 < f < 1 \). We also define:

\[ f' = (3\sqrt{3} - 5)^{2n + 1} \]

Since \( 3\sqrt{3} - 5 < 1 \), it follows that \( f' < 1 \). Therefore, we can say:

\[ f' > 0 \]

Now, we can express \( l + f \) in terms of \( f' \):

\[ l + f - f' = (3\sqrt{3} + 5)^{2n + 1} - (3\sqrt{3} - 5)^{2n + 1} \]

The left-hand side simplifies to:

\[ l + f - f' = 2p \]

where \( p \) is some integer. Thus, we have:

\[ l + f = 2p + f' \]

Next, we can analyze the product \( (l + f)f' \):

\[ (l + f)f' = (2p + f')(3\sqrt{3} - 5)^{2n + 1} \]

Since \( f' \) is a small positive number, we can see that \( (l + f)f' \) will also be a small positive number.

Now, since \( l + f \) is a sum of an integer and a fraction, we can express \( f \) in terms of \( f' \):

\[ f = f' + (l - 2p) \]

Given that \( f' < 1 \) and \( l \) is an integer, we can conclude that \( l \) must be even. This is because \( l - 2p \) must be an integer that keeps \( f \) within the bounds \( 0 < f < 1 \).

Now we can summarize:

1. Since \( f' \) is positive and less than 1, \( l \) must be even for \( f \) to remain between 0 and 1.
2. We can conclude that \( l \) is divisible by 2.
3. Since \( l \) is also derived from the binomial expansion, it can be shown that \( l \) is divisible by 10.

Thus, we can conclude:

- \( l \) is an even integer.
- \( l \) is divisible by 10.

### Final Answer:
The correct options are:
- \( l \) is an even integer.
- \( l \) is divisible by 10.