Let `x=(8sqrt(3)+13)^(13)` and `y=(7sqrt(2)+9)^(9)`.If [t] denotes the greatest integer `let`,then

Options

A. `[x]` is even but `[y]` is odd
B. `[x]+[y]` is even
C. ` [x] and [y] are both odd
D. `[x]` is odd but `[y]` is even

Answer

To solve the problem, we need to evaluate the expressions for \( x \) and \( y \) and then find the greatest integer less than or equal to \( x + y \).

### Step 1: Evaluate \( x = (8\sqrt{3} + 13)^{13} \)

We can express \( x \) in terms of its integer and fractional parts:
\[ x = (8\sqrt{3} + 13)^{13} = I + F \]
where \( I \) is the integer part and \( F \) is the fractional part.

### Step 2: Find the approximate value of \( 8\sqrt{3} + 13 \)

Calculating \( 8\sqrt{3} \):
\[ \sqrt{3} \approx 1.732 \implies 8\sqrt{3} \approx 8 \times 1.732 \approx 13.856 \]
Thus,
\[ 8\sqrt{3} + 13 \approx 13.856 + 13 = 26.856 \]

### Step 3: Raise to the power of 13

Now, we need to evaluate \( (26.856)^{13} \). To simplify, we can use the binomial theorem:
\[ (8\sqrt{3} + 13)^{13} = \sum_{k=0}^{13} \binom{13}{k} (8\sqrt{3})^k (13)^{13-k} \]

### Step 4: Approximate the leading term

The leading term (when \( k = 0 \)) is:
\[ 13^{13} \]
The next term (when \( k = 1 \)) is:
\[ \binom{13}{1} (8\sqrt{3})^1 (13)^{12} = 13 \cdot (8\sqrt{3}) \cdot 13^{12} = 13^{12} \cdot 8\sqrt{3} \cdot 13 \]

### Step 5: Find \( F \)

Since \( 8\sqrt{3} < 13 \), we can say:
\[ (8\sqrt{3} - 13)^{13} \text{ is a small positive number.} \]
Thus, \( F \) is a small positive number, and we can conclude:
\[ I = \lfloor (8\sqrt{3} + 13)^{13} \rfloor = 13^{13} + \text{small positive number} \]

### Step 6: Evaluate \( y = (7\sqrt{2} + 9)^{9} \)

Similarly, we can express \( y \) as:
\[ y = (7\sqrt{2} + 9)^{9} = I' + F' \]
where \( I' \) is the integer part and \( F' \) is the fractional part.

### Step 7: Find the approximate value of \( 7\sqrt{2} + 9 \)

Calculating \( 7\sqrt{2} \):
\[ \sqrt{2} \approx 1.414 \implies 7\sqrt{2} \approx 7 \times 1.414 \approx 9.898 \]
Thus,
\[ 7\sqrt{2} + 9 \approx 9.898 + 9 = 18.898 \]

### Step 8: Raise to the power of 9

Using the binomial theorem:
\[ (7\sqrt{2} + 9)^{9} = \sum_{k=0}^{9} \binom{9}{k} (7\sqrt{2})^k (9)^{9-k} \]
The leading term is:
\[ 9^{9} \]
The next term is:
\[ \binom{9}{1} (7\sqrt{2})^1 (9)^{8} = 9 \cdot (7\sqrt{2}) \cdot 9^{8} \]

### Step 9: Find \( F' \)

Since \( 7\sqrt{2} < 9 \), we can conclude:
\[ (7\sqrt{2} - 9)^{9} \text{ is a small positive number.} \]
Thus, \( I' = \lfloor (7\sqrt{2} + 9)^{9} \rfloor = 9^{9} + \text{small positive number} \)

### Step 10: Combine \( x \) and \( y \)

Now we can combine \( x \) and \( y \):
\[ \lfloor x + y \rfloor = \lfloor I + F + I' + F' \rfloor = I + I' + \lfloor F + F' \rfloor \]

Since both \( F \) and \( F' \) are small positive numbers, we can conclude:
\[ \lfloor F + F' \rfloor = 0 \]

### Final Answer

Thus, the greatest integer less than or equal to \( x + y \) is:
\[ \lfloor x + y \rfloor = I + I' = 13^{13} + 9^{9} \]