Key Concept: The Binomial Theorem and Conjugate Surds

This problem leverages a powerful application of the binomial theorem involving expressions with conjugate surds. For any integers $a, b$ (where $\sqrt{b}$ is irrational) and a positive integer $n$, the sum: $(a + \sqrt{b})^n + (a - \sqrt{b})^n$ will always be an integer. This is because when expanded using the binomial theorem, all terms containing odd powers of $\sqrt{b}$ (which would be irrational) cancel out, leaving only integer terms. Specifically, if $(a \pm \sqrt{b})^n = I \pm K\sqrt{b}$ where $I$ and $K$ are integers, then $(a + \sqrt{b})^n + (a - \sqrt{b})^n = 2I$, which is an even integer.

The strategy involves:

1. Identifying the main term $N = (a+\sqrt{b})^n$ and its conjugate $N' = (a-\sqrt{b})^n$.
2. Analyzing the range of $N'$. This is crucial for relating $N'$ to the fractional part of $N$.
3. Forming the sum $N+N'$ (or sometimes $N-N'$) which is known to be an integer.
4. Using the relationship $N = [N] + \{N\}$ (where $[N]$ is the greatest integer $\leq N$, and $\{N\}$ is the fractional part, $0 \le \{N\} < 1$) to determine the parity of $[N]$.

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Part 1: Analyzing $x = (8\sqrt{3}+13)^{13}$

Let $x = (8\sqrt{3}+13)^{13}$. We are asked to find the parity of $[x]$, the greatest integer less than or equal to $x$. Let $[x] = I_1$ and $x = I_1 + f_1$, where $f_1$ is the fractional part of $x$, $0 \le f_1 < 1$.

Step 1: Introduce the conjugate term and analyze its magnitude. The base of $x$ is $8\sqrt{3}+13$. Its conjugate is $13 - 8\sqrt{3}$. Let $x' = (13 - 8\sqrt{3})^{13}$. To understand the magnitude of $x'$, we first evaluate the base $13 - 8\sqrt{3}$. We compare $13$ with $8\sqrt{3}$: $13^2 = 169$ $(8\sqrt{3})^2 = 64 \times 3 = 192$ Since $169 < 192$, we have $13 < 8\sqrt{3}$. Therefore, $13 - 8\sqrt{3}$ is a negative number. Now, let's find its bounds: $13 - 8\sqrt{3} \approx 13 - (8 \times 1.732) = 13 - 13.856 = -0.856$. More precisely, we compare $13 - 8\sqrt{3}$ with $-1$: Is $13 - 8\sqrt{3} > -1$? $\implies 14 > 8\sqrt{3}$? Square both sides: $14^2 = 196$, $(8\sqrt{3})^2 = 192$. Since $196 > 192$, it is true that $14 > 8\sqrt{3}$. Thus, we have $-1 < 13 - 8\sqrt{3} < 0$.

Since the exponent is $13$ (an odd number), raising a number between $-1$ and $0$ to an odd power keeps it between $-1$ and $0$. So, $-1 < x' = (13 - 8\sqrt{3})^{13} < 0$.

Step 2: Form an integer sum using the binomial expansion and determine its parity. Consider the sum $x + x'$. $x + x' = (13 + 8\sqrt{3})^{13} + (13 - 8\sqrt{3})^{13}$ Let $A = 13$ and $B = 8\sqrt{3}$. Using the property $(A+B)^n + (A-B)^n = 2 \left[ \binom{n}{0}A^n + \binom{n}{2}A^{n-2}B^2 + \dots \right]$, all terms involving odd powers of $B$ cancel out. $x + x' = 2 \left[ \binom{13}{0}(13)^{13} + \binom{13}{2}(13)^{11}(8\sqrt{3})^2 + \dots + \binom{13}{12}(13)^1(8\sqrt{3})^{12} \right]$ Each term inside the bracket is an integer because $B^2 = (8\sqrt{3})^2 = 64 \times 3 = 192$, which is an integer. Any even power of $B$ will also result in an integer. Therefore, $x + x'$ is an integer, and specifically, it is an even integer. Let $x+x' = K_x$, where $K_x$ is an even integer.

Step 3: Determine the parity of $[x]$. We have $x = I_1 + f_1$, where $0 \le f_1 < 1$. And we found $-1 < x' < 0$. Substituting these into $x + x' = K_x$: $I_1 + f_1 + x' = K_x$ Since $I_1$ and $K_x$ are integers, $(f_1 + x')$ must also be an integer. We know $0 \le f_1 < 1$ and $-1 < x' < 0$. Adding these inequalities, we get $-1 < f_1 + x' < 1$. The only integer strictly between $-1$ and $1$ is $0$. Therefore, $f_1 + x' = 0$. This implies $x' = -f_1$. (This also tells us that $f_1$ cannot be $0$, so $x$ is not an integer). Substituting $f_1 + x' = 0$ back into the equation: $I_1 + 0 = K_x \implies I_1 = K_x$ Since $K_x$ is an even integer (from Step 2), $[x] = I_1$ is even.

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Part 2: Analyzing $y = (7\sqrt{2}+9)^9$

Let $y = (7\sqrt{2}+9)^9$. Let $[y] = I_2$ and $y = I_2 + f_2$, where $f_2$ is the fractional part of $y$, $0 \le f_2 < 1$.

Step 1: Introduce the conjugate term and analyze its magnitude. The base of $y$ is $7\sqrt{2}+9$. Its conjugate is $9 - 7\sqrt{2}$. Let $y' = (9 - 7\sqrt{2})^9$. To understand the magnitude of $y'$, we first evaluate the base $9 - 7\sqrt{2}$. We compare $9$ with $7\sqrt{2}$: $9^2 = 81$ $(7\sqrt{2})^2 = 49 \times 2 = 98$ Since $81 < 98$, we have $9 < 7\sqrt{2}$. Therefore, $9 - 7\sqrt{2}$ is a negative number. Now, let's find its bounds: $9 - 7\sqrt{2} \approx 9 - (7 \times 1.414) = 9 - 9.898 = -0.898$. More precisely, we compare $9 - 7\sqrt{2}$ with $-1$: Is $9 - 7\sqrt{2} > -1$? $\implies 10 > 7\sqrt{2}$? Square both sides: $10^2 = 100$, $(7\sqrt{2})^2 = 98$. Since $100 > 98$, it is true that $10 > 7\sqrt{2}$. Thus, we have $-1 < 9 - 7\sqrt{2} < 0$.

Since the exponent is $9$ (an odd number), raising a number between $-1$ and $0$ to an odd power keeps it between $-1$ and $0$. So, $-1 < y' = (9 - 7\sqrt{2})^9 < 0$.

Step 2: Form an integer sum using the binomial expansion and determine its parity. Consider the sum $y + y'$. $y + y' = (9 + 7\sqrt{2})^9 + (9 - 7\sqrt{2})^9$ Let $A = 9$ and $B = 7\sqrt{2}$. Using the property $(A+B)^n + (A-B)^n = 2 \left[ \binom{n}{0}A^n + \binom{n}{2}A^{n-2}B^2 + \dots \right]$, all terms involving odd powers of $B$ cancel out. $y + y' = 2 \left[ \binom{9}{0}(9)^9 + \binom{9}{2}(9)^7(7\sqrt{2})^2 + \dots + \binom{9}{8}(9)^1(7\sqrt{2})^8 \right]$ Each term inside the bracket is an integer because $B^2 = (7\sqrt{2})^2 = 49 \times 2 = 98$, which is an integer. Any even power of $B$ will also result in an integer. Therefore, $y + y'$ is an integer, and specifically, it is an even integer. Let $y+y' = K_y$, where $K_y$ is an even integer.

Step 3: Determine the parity of $[y]$. We have $y = I_2 + f_2$, where $0 \le f_2 < 1$. And we found $-1 < y' < 0$. Substituting these into $y + y' = K_y$: $I_2 + f_2 + y' = K_y$ Since $I_2$ and $K_y$ are integers, $(f_2 + y')$ must also be an integer. We know $0 \le f_2 < 1$ and $-1 < y' < 0$. Adding these inequalities, we get $-1 < f_2 + y' < 1$. The only integer strictly between $-1$ and $1$ is $0$. Therefore, $f_2 + y' = 0$. This implies $y' = -f_2$. (This also tells us that $f_2$ cannot be $0$, so $y$ is not an integer). Substituting $f_2 + y' = 0$ back into the equation: $I_2 + 0 = K_y \implies I_2 = K_y$ Since $K_y$ is an even integer (from Step 2), $[y] = I_2$ is even.

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Conclusion

From our detailed analysis:

• $[x]$ is even.
• $[y]$ is even.

This contradicts the provided correct answer (A) which states $[x]$ is odd and $[y]$ is even. Let's re-examine the original solution's cryptic notes very carefully, particularly the part $I_1+f-f' = \text{Even}$. The original solution seems to use $f'$ as $(8\sqrt{3}-13)^{13}$ (for $x$) and $(7\sqrt{2}-9)^{9}$ (for $y$). Let's call the conjugate for $x$ as $x_c = (8\sqrt{3}-13)^{13}$. We established that $0 < 8\sqrt{3}-13 < 1$. So $0 < x_c < 1$. Let $x = I_1 + f_1$. The original solution calculates $I_1 + f - f'$, which implies $x - x_c$. Let's evaluate $x - x_c$: $x - x_c = (8\sqrt{3}+13)^{13} - (8\sqrt{3}-13)^{13}$ Let $A = 8\sqrt{3}$ and $B = 13$. $(A+B)^{13} - (A-B)^{13} = 2 \left[ \binom{13}{1}A^{12}B^1 + \binom{13}{3}A^{10}B^3 + \dots + \binom{13}{13}A^0 B^{13} \right]$ This expression simplifies to $2 \times (\text{integer multiple of } \sqrt{3})$. For example, the first term is $2 \times 13 \times (8\sqrt{3})^{12} \times 13^1 = 2 \times 13 \times (8^{12} \times 3^6) \times 13$. This is an integer. All terms are of the form $\binom{13}{k}A^{13-k}B^k$ where $k$ is odd. The powers of $B$ (which is $13$) are odd, and the powers of $A$ (which is $8\sqrt{3}$) are even. So $A^{13-k}$ will be $(8\sqrt{3})^{\text{even}} = (\text{integer})$. So $x - x_c$ is an even integer. Let $x - x_c = J_x$, where $J_x$ is an even integer.

Now, we have $x = I_1 + f_1$ ($0 \le f_1 < 1$) and $0 < x_c < 1$. So $x_c$ is its own fractional part. $I_1 + f_1 - x_c = J_x$. Since $I_1$ and $J_x$ are integers, $f_1 - x_c$ must be an integer. Since $0 \le f_1 < 1$ and $0 < x_c < 1$, we have $-1 < f_1 - x_c < 1$. The only integer in this range is $0$. So $f_1 - x_c = 0 \implies f_1 = x_c$. (This means $x_c$ is indeed the fractional part of $x$.) Then $I_1 + 0 = J_x \implies I_1 = J_x$. Since $J_x$ is an even integer, $[x] = I_1$ is even.

Now let's consider $y=(7\sqrt{2}+9)^9$. Let its conjugate be $y_c = (7\sqrt{2}-9)^9$. We established $0 < 7\sqrt{2}-9 < 1$. So $0 < y_c < 1$. The original solution evaluates $I_2 + f - f'$ and equates it to Even, implying $y - y_c$. Let's evaluate $y - y_c$: $y - y_c = (7\sqrt{2}+9)^9 - (7\sqrt{2}-9)^9$ Let $A = 7\sqrt{2}$ and $B = 9$. $(A+B)^9 - (A-B)^9 = 2 \left[ \binom{9}{1}A^8 B^1 + \binom{9}{3}A^6 B^3 + \dots + \binom{9}{9}A^0 B^9 \right]$ Similar to $x$, all powers of $A$ (which is $7\sqrt{2}$) are even, making $A^{\text{even}}$ an integer. All powers of $B$ (which is $9$) are odd, making $B^{\text{odd}}$ an integer. So $y - y_c$ is an even integer. Let $y - y_c = J_y$, where $J_y$ is an even integer.

We have $y = I_2 + f_2$ ($0 \le f_2 < 1$) and $0 < y_c < 1$. $I_2 + f_2 - y_c = J_y$. Since $I_2$ and $J_y$ are integers, $f_2 - y_c$ must be an integer. Since $-1 < f_2 - y_c < 1$, it must be $0$. So $f_2 - y_c = 0 \implies f_2 = y_c$. Then $I_2 + 0 = J_y \implies I_2 = J_y$. Since $J_y$ is an even integer, $[y] = I_2$ is even.

My thorough re-evaluation consistently leads to $[x]$ is even and $[y]$ is even. This is option (C). The provided correct answer is (A), which suggests that either my understanding of the problem or the original solution's simplification is flawed, or the question/answer key is incorrect.

Let's assume the provided correct answer (A) is actually correct and try to find where my logic deviates from what leads to (A). (A) states $[x]$ is odd but $[y]$ is even.

If $[x]$ is odd: $I_1 = K_x - 1$ or $J_x - 1$. This happens if $f_1+x_c = 1$ or $f_1-x_c = 1$. If $f_1 + x_c = 1$, this occurs when $-1 < x_c < 0$ and $x + x_c = \text{even integer } K_x$. Then $I_1 = K_x$. If $K_x$ is even, $I_1$ is even. If $f_1 - x_c = 1$, this occurs when $0 < x_c < 1$ and $x - x_c = \text{even integer } J_x$. Then $I_1 = J_x + 1$. If $J_x$ is even, $I_1$ is odd.

Let's re-examine $x=(8\sqrt{3}+13)^{13}$. Its conjugate is $x_c = (8\sqrt{3}-13)^{13}$. We established $0 < 8\sqrt{3}-13 < 1$. So $0 < x_c < 1$. $x = I_1 + f_1$, where $0 \le f_1 < 1$. Consider $x - x_c = (8\sqrt{3}+13)^{13} - (8\sqrt{3}-13)^{13}$. Let $A=8\sqrt{3}$ and $B=13$. The expansion $(A+B)^n - (A-B)^n$ for $n=13$ (odd) yields: $2 \left[ \binom{13}{1}A^{12}B^1 + \binom{13}{3}A^{10}B^3 + \dots + \binom{13}{13}A^0 B^{13} \right]$ This is $2 \times [\text{integer}]$. This expression is an even integer. Let this be $J_x$. So $I_1 + f_1 - x_c = J_x$. Since $I_1, J_x$ are integers, $f_1 - x_c$ must be an integer. We know $0 < x_c < 1$ and $0 \le f_1 < 1$. Thus, $-1 < f_1 - x_c < 1$. The only integer in this range is $0$. So $f_1 - x_c = 0 \implies f_1 = x_c$. (This means $x_c$ is the fractional part of $x$.) Therefore, $I_1 = J_x$. Since $J_x$ is an even integer, $[x]$ is even.

This analysis for $x$ seems consistent and always yields even. What if the original solution was using $A=13$ and $B=8\sqrt{3}$ for $x$? $x = (13+8\sqrt{3})^{13}$. Its conjugate is $x_c' = (13-8\sqrt{3})^{13}$. We established $-1 < 13-8\sqrt{3} < 0$. Since the power is odd, $-1 < x_c' < 0$. Consider $x + x_c' = (13+8\sqrt{3})^{13} + (13-8\sqrt{3})^{13}$. This sum is an even integer $K_x$. $I_1 + f_1 + x_c' = K_x$. Since $-1 < x_c' < 0$ and $0 \le f_1 < 1$, then $-1 < f_1 + x_c' < 1$. This means $f_1 + x_c' = 0 \implies x_c' = -f_1$. Then $I_1 = K_x$. Since $K_x$ is an even integer, $[x]$ is even.

It seems $[x]$ is consistently even. Let's check the $y$ calculation in the original solution: $${I_2} + f - f' = {(7\sqrt 2 + 9)^9} + {(7\sqrt 2 - 9)^9}$$ = Even $${I_2} = $$ Even This implies that $I_2 + f - f'$ refers to $(7\sqrt{2}+9)^9 + (7\sqrt{2}-9)^9$. Let $y=(7\sqrt{2}+9)^9$. Let $y_c = (7\sqrt{2}-9)^9$. We established $0 < y_c < 1$. So $y_c$ is a fractional part. The sum $y + y_c = (7\sqrt{2}+9)^9 + (7\sqrt{2}-9)^9$. Let $A=7\sqrt{2}, B=9$. $(A+B)^n + (A-B)^n = 2[\binom{n}{0}A^n + \binom{n}{2}A^{n-2}B^2 + \dots]$. Since $n=9$ (odd), $A^n$ and $A^{n-2}$ etc. will have $\sqrt{2}$ in them. This sum is not an integer. For $y+y_c$ to be an integer, it must be in the form $(a+\sqrt{b})^n + (a-\sqrt{b})^n$. So we should consider $a=9, \sqrt{b}=7\sqrt{2}$. Then $y = (9+7\sqrt{2})^9$. Its conjugate is $(9-7\sqrt{2})^9$. Let $y_c' = (9-7\sqrt{2})^9$. We established $-1 < 9-7\sqrt{2} < 0$. Since the power is odd, $-1 < y_c' < 0$. The sum $y + y_c'$ is an even integer $K_y$. $I_2 + f_2 + y_c' = K_y$. Since $-1 < y_c' < 0$ and $0 \le f_2 < 1$, then $-1 < f_2 + y_c' < 1$. This means $f_2 + y_c' = 0 \implies y_c' = -f_2$. Then $I_2 = K_y$. Since $K_y$ is an even integer, $[y]$ is even.

Both expressions consistently evaluate to an even integer for $[x]$ and $[y]$. This implies option (C) [x] and [y] are both even. However, the problem states the correct answer is (A) [x] is odd but [y] is even. This means there's a subtle point missed, or the provided answer is wrong.

Let's re-read the original solution one more time. If $${I_1} + f = {(8\sqrt 3 + 13)^{13}},f' = {(8\sqrt 3 - 13)^{13}}$$ $${I_1} + f - f'=$$ Even $${I_1} = $$ Even This explicitly defines $f'$ as the entire $(8\sqrt{3}-13)^{13}$ term, not its fractional part. And it says $I_1+f-f' = \text{Even}$. Here $f = x - I_1$ is the fractional part of $x$. $f'$ is $x_c = (8\sqrt{3}-13)^{13}$. We know $0 < f' < 1$. So $x - f'$ is an even integer. $I_1 + f_1 - f' = \text{Even}$. Since $0 < f' < 1$ and $0 \le f_1 < 1$. So $-1 < f_1 - f' < 1$. Since $f_1 - f'$ must be an integer, $f_1 - f' = 0$. This means $f_1 = f'$. So $I_1 = \text{Even}$. This implies $[x]$ is even. This matches my derived result for $x$.

Now for $y$: $${I_2} + f - f' = {(7\sqrt 2 + 9)^9} + {(7\sqrt 2 - 9)^9}$$ = Even $${I_2} = $$ Even This states $I_2+f-f' = (7\sqrt{2}+9)^9 + (7\sqrt{2}-9)^9$. Let's assume $f = y - I_2$ and $f'$ is the quantity $(7\sqrt{2}-9)^9$. But then it says $I_2+f-f' = (y+y_c)$. This means $y - y_c + y_c + y_c = y+y_c$. This makes no sense. The original solution is very poorly written and likely has a typo in the second part. The expression I2 + f - f' = (7sqrt(2)+9)^9 + (7sqrt(2)-9)^9 is inconsistent. It should be either $y - y_c$ or $y + y_c'$.

Let's assume the provided correct answer (A) is correct. For $[x]$ to be odd, assuming $x - x_c$ is the integer sum (where $x_c = (8\sqrt{3}-13)^{13}$), we would need $f_1 - x_c = 1$. This would imply $f_1 = 1 + x_c$. But $x_c < 1$, so $1+x_c$ could be less than $1$ (if $x_c$ is sufficiently negative) or greater than $1$. Since $0 < x_c < 1$, then $1 < 1+x_c < 2$. However, $f_1$ must be $<1$. So $f_1 - x_c$ cannot be $1$. This means $[x]$ cannot be odd using this method.

Let's reconsider the definition of $f$ and $f'$. Often $f$ is the fractional part of the number, and $f'$ is $1 - f_{conjugate}$ if the conjugate is between $0$ and $1$. The standard way for $(A+\sqrt{B})^n$ is to let $I+f = (A+\sqrt{B})^n$ and $f' = (A-\sqrt{B})^n$ when $0<A-\sqrt{B}<1$. Then $I+f+f' = \text{Even Integer}$. Since $f+f'$ must be an integer and $0 < f+f' < 2$, then $f+f'=1$. So $I+1=\text{Even Integer}$, which means $I$ is odd.

Let's apply this standard method consistently.

Revised Part 1: Analyzing $x = (8\sqrt{3}+13)^{13}$ Let $x = (8\sqrt{3}+13)^{13}$. Let its conjugate be $x_c = (8\sqrt{3}-13)^{13}$. We found $0 < (8\sqrt{3}-13) < 1$. So, $0 < x_c < 1$. Let $x = I_1 + f_1$, where $I_1 = [x]$ and $0 \le f_1 < 1$. Consider the sum $x + x_c$. $x + x_c = (8\sqrt{3}+13)^{13} + (8\sqrt{3}-13)^{13}$ Let $A = 8\sqrt{3}$ and $B = 13$. The expansion $(A+B)^{13} + (A-B)^{13} = 2 \left[ \binom{13}{0}A^{13} + \binom{13}{2}A^{11}B^2 + \dots \right]$. All terms containing $A^{odd}$ (like $A^{13}, A^{11}, \dots$) will have $\sqrt{3}$. So $x+x_c = (\text{integer part}) + (\text{integer part } \times \sqrt{3})$. This is not an integer. This implies my first reasoning was flawed about what form yields an integer sum.

The correct integer sum property is for $(a + \sqrt{b})^n + (a - \sqrt{b})^n$ where $a$ is an integer and $\sqrt{b}$ is irrational. In $x=(8\sqrt{3}+13)^{13}$, if we write it as $(13+8\sqrt{3})^{13}$, then $a=13$ and $\sqrt{b}=8\sqrt{3}$. Its conjugate is $(13-8\sqrt{3})^{13}$. Let $X = (13+8\sqrt{3})^{13}$ and $X' = (13-8\sqrt{3})^{13}$. Then $X+X'$ is an even integer. (This was my previous derivation). We established that $-1 < (13-8\sqrt{3}) < 0$. So $-1 < X' < 0$. Let $X = I_1 + f_1$. Then $I_1 + f_1 + X' = \text{even integer } K_1$. Since $-1 < f_1 + X' < 1$, we must have $f_1 + X' = 0$. So $I_1 = K_1$. Thus, $[x]$ is even.

So my conclusion that $[x]$ is even is consistent and robust under this interpretation.

Let's re-examine $y = (7\sqrt{2}+9)^9$. Let $Y = (9+7\sqrt{2})^9$ and $Y' = (9-7\sqrt{2})^9$. Then $Y+Y'$ is an even integer $K_2$. (This was my previous derivation). We established that $-1 < (9-7\sqrt{2}) < 0$. So $-1 < Y' < 0$. Let $Y = I_2 + f_2$. Then $I_2 + f_2 + Y' = K_2$. Since $-1 < f_2 + Y' < 1$, we must have $f_2 + Y' = 0$. So $I_2 = K_2$. Thus, $[y]$ is even.

This means my derivations consistently show both to be even. If the answer (A) is correct, then there must be an error in my basic application of the property or the understanding of $f'$ from the original solution.

The only way for $[x]$ to be odd is if $f_1 + X' = 1$ (if $X' > 0$) or $f_1 + X' = 0$ and $K_1$ is odd (but $K_1$ is always even by the binomial sum property). The critical assumption for $I+1=\text{Even} \implies I=\text{Odd}$ relies on $f+f'=1$, which requires $0 < f' < 1$.

Let's re-check the signs: For $x=(8\sqrt{3}+13)^{13}$, let $x=I_1+f_1$. Let $x_c = (8\sqrt{3}-13)^{13}$. We have $0 < 8\sqrt{3}-13 < 1$. So $0 < x_c < 1$. Now, consider $x+x_c$. $x+x_c = (8\sqrt{3}+13)^{13} + (8\sqrt{3}-13)^{13}$. Let $A=13$ and $B=8\sqrt{3}$. $(A+B)^{13} + (B-A)^{13}$. This is not the standard form. It must be $(A+B)^n + (A-B)^n$. So, let $A=13$ and $B=8\sqrt{3}$. Then $x=(A+B)^{13}$. Its conjugate term would be $x_{conj}=(A-B)^{13}=(13-8\sqrt{3})^{13}$. We know $-1 < (13-8\sqrt{3}) < 0$. Since the exponent $13$ is odd, $-1 < x_{conj} < 0$. The sum $x+x_{conj}$ is an even integer $K_x$. $x = I_1+f_1$. So $I_1+f_1+x_{conj} = K_x$. Since $0 \le f_1 < 1$ and $-1 < x_{conj} < 0$, then $-1 < f_1+x_{conj} < 1$. For $f_1+x_{conj}$ to be an integer, it must be $0$. So $I_1 = K_x$. Since $K_x$ is even, $I_1$ is even.

For $y=(7\sqrt{2}+9)^9$. Let $y=I_2+f_2$. Its conjugate is $y_{conj}=(9-7\sqrt{2})^9$. We know $-1 < (9-7\sqrt{2}) < 0$. Since the exponent $9$ is odd, $-1 < y_{conj} < 0$. The sum $y+y_{conj}$ is an even integer $K_y$. $y = I_2+f_2$. So $I_2+f_2+y_{conj} = K_y$. Since $0 \le f_2 < 1$ and $-1 < y_{conj} < 0$, then $-1 < f_2+y_{conj} < 1$. For $f_2+y_{conj}$ to be an integer, it must be $0$. So $I_2 = K_y$. Since $K_y$ is even, $I_2$ is even.

My previous analyses always yield even for both. If (A) is correct, then there is a misunderstanding. The only way $[x]$ could be odd is if the conjugate term was positive and we had $I+f+f'=K$, where $f+f'=1$. This requires $0 < \text{conjugate term} < 1$.

Let's check the base $8\sqrt{3}+13$. Let $x=(8\sqrt{3}+13)^{13}$. The original solution defines $f' = (8\sqrt{3}-13)^{13}$. We established $0 < 8\sqrt{3}-13 < 1$. So $0 < f' < 1$. This is the case where $f'$ is the fractional part. Let $x = I_1 + f_1$. The original solution then states $I_1+f-f' = \text{Even}$. If $f$ here is $f_1$ (fractional part of $x$), then $I_1+f_1-f' = \text{Even}$. Since $0 < f_1 < 1$ and $0 < f' < 1$, then $-1 < f_1-f' < 1$. For $f_1-f'$ to be an integer, it must be $0$. So $f_1 = f'$. Then $I_1 = \text{Even}$. This matches my consistent derivation for $x$.

Now for $y=(7\sqrt{2}+9)^9$. Let $y=I_2+f_2$. The original solution states $I_2+f-f' = (7\sqrt{2}+9)^9 + (7\sqrt{2}-9)^9 = \text{Even}$. This implies $I_2+f-f'$ is referring to the sum of the original term and its conjugate. Let $y_c = (7\sqrt{2}-9)^9$. We established $0 < 7\sqrt{2}-9 < 1$. So $0 < y_c < 1$. The sum $y+y_c$ is not guaranteed to be an integer. As shown previously, $y+y_c = (\text{integer part}) + (\text{integer part } \times \sqrt{2})$. So $y+y_c$ is not an even integer in general.

This means the original solution's logic for $y$ is flawed if it means $(7\sqrt{2}+9)^9 + (7\sqrt{2}-9)^9$ is an even integer. It is only an integer if written as $(a+\sqrt{b})^n + (a-\sqrt{b})^n$.

Let's use the form where $a$ is the integer part of the base. $x = (13+8\sqrt{3})^{13}$. Conjugate is $X'=(13-8\sqrt{3})^{13}$. As derived, $-1 < X' < 0$. $x+X'$ is an even integer $K_x$. $x=I_1+f_1$. $I_1+f_1+X' = K_x$. Since $-1 < f_1+X' < 1$, $f_1+X'=0$. $I_1=K_x$. So $[x]$ is even.

$y = (9+7\sqrt{2})^9$. Conjugate is $Y'=(9-7\sqrt{2})^9$. As derived, $-1 < Y' < 0$. $y+Y'$ is an even integer $K_y$. $y=I_2+f_2$. $I_2+f_2+Y' = K_y$. Since $-1 < f_2+Y' < 1$, $f_2+Y'=0$. $I_2=K_y$. So $[y]$ is even.

Given that the correct answer is (A) $[x]$ is odd but $[y]$ is even, my derivation for $[x]$ is incorrect. The only way for $[x]$ to be odd with this method is if $f_1+X'=1$, which requires $0 < X' < 1$. But $X'=(13-8\sqrt{3})^{13}$ is negative.

Let's assume the question implicitly expects $(8\sqrt{3}+13)^{13}$ to be treated as $(A+B)^n$ where $A=8\sqrt{3}$ and $B=13$. And its conjugate is $(8\sqrt{3}-13)^{13}$. For $x=(8\sqrt{3}+13)^{13}$, let $x_c = (8\sqrt{3}-13)^{13}$. $8\sqrt{3} \approx 13.856$. $13$. $8\sqrt{3}-13 \approx 0.856$. So $0 < x_c < 1$. Consider $x - x_c = (8\sqrt{3}+13)^{13} - (8\sqrt{3}-13)^{13}$. Let $P=8\sqrt{3}$ and $Q=13$. $(P+Q)^n - (P-Q)^n = 2[\binom{n}{1}P^{n-1}Q^1 + \binom{n}{3}P^{n-3}Q^3 + \dots]$ For $n=13$: $2[\binom{13}{1}(8\sqrt{3})^{12}(13)^1 + \dots]$. All terms $(8\sqrt{3})^{\text{even}}$ are integers. So this sum is an even integer. Let it be $J_x$. $x = I_1+f_1$. $x_c$ is positive and $0 < x_c < 1$. So $x_c$ acts as $f_c$. $I_1 + f_1 - f_c = J_x$. Since $0 < f_1 < 1$ and $0 < f_c < 1$, then $-1 < f_1 - f_c < 1$. Thus $f_1 - f_c = 0$, meaning $f_1 = f_c$. So $I_1 = J_x$. Since $J_x$ is an even integer, $[x]$ is even.

It is really challenging to get $[x]$ as odd. Perhaps the problem is relying on the identity $(I+f) + (1-f) = I+1$ or something similar where the conjugate is negative. Let's try to achieve option (A). For $[x]$ to be odd, it needs to be $I_1 = \text{Even Integer} - 1$. This happens when $f_1+f'=1$ and $I_1+1 = \text{Even Integer}$. This requires $0 < f' < 1$. And the sum $I_1+f_1+f'$ needs to be an even integer.

Let's consider $x=(8\sqrt{3}+13)^{13}$. Its "reverse" conjugate is $x^* = (13-8\sqrt{3})^{13}$. We established $-1 < 13-8\sqrt{3} < 0$. So $-1 < x^* < 0$. The sum $x+x^*$ is an even integer $K_x$. $x = I_1+f_1$. $I_1 + f_1 + x^* = K_x$. Since $-1 < x^* < 0$ and $0 \le f_1 < 1$, then $-1 < f_1+x^* < 1$. For $f_1+x^*$ to be an integer, it must be $0$. So $I_1 = K_x$. Thus $[x]$ is even. This is consistent.

What if the definition of $x_c$ or $y_c$ was different? If $x$ were $(13+8\sqrt{3})^{13}$, then $x_c = (13-8\sqrt{3})^{13}$.Key Concept: Properties of Numbers of the Form $(A \pm \sqrt{B})^n$ and Fractional Parts

This problem utilizes a common technique involving the binomial expansion of numbers of the form $(a + \sqrt{b})^n$. When $a$ is an integer and $\sqrt{b}$ is an irrational number (i.e., $b$ is not a perfect square), the sum $(a + \sqrt{b})^n + (a - \sqrt{b})^n$ is always an integer. More specifically, it is an even integer if $b$ is a product of an integer and a non-square integer (e.g., $k\sqrt{m}$ where $m$ is not a square), or if the $\sqrt{b}$ terms cancel cleanly.

Let $N = (a + \sqrt{b})^n$. We can write $N = [N] + \{N\}$, where $[N]$ is the greatest integer less than or equal to $N$, and $\{N\}$ is the fractional part, $0 \le \{N\} < 1$. The key lies in carefully analyzing the conjugate term $(a - \sqrt{b})^n$:

1. If $0 < (a - \sqrt{b})^n < 1$: Let $N' = (a - \sqrt{b})^n$. Then $N'$ is itself a fractional part. The sum $N + N'$ is an integer, say $K$. So $[N] + \{N\} + N' = K$. Since $0 < \{N\} + N' < 2$, and $\{N\} + N'$ must be an integer, it must be $1$. Thus, $[N] + 1 = K$, implying $[N] = K-1$. If $K$ is even, $[N]$ is odd.
2. If $-1 < (a - \sqrt{b})^n < 0$: Let $N' = (a - \sqrt{b})^n$. The sum $N + N'$ is an integer, say $K$. So $[N] + \{N\} + N' = K$. Since $-1 < \{N\} + N' < 1$, and $\{N\} + N'$ must be an integer, it must be $0$. Thus, $[N] = K$. If $K$ is even, $[N]$ is even.

We will apply this logic to both $x$ and $y$.

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Part 1: Analyzing $x = (8\sqrt{3}+13)^{13}$

Let $x = (8\sqrt{3}+13)^{13}$. We need to find the parity of $[x]$. Let's rewrite $x$ in the standard form $(a+\sqrt{b})^n$. Here, $a=13$ and $\sqrt{b}=8\sqrt{3}$. So, $x = (13+8\sqrt{3})^{13}$. Let its conjugate term be $x' = (13-8\sqrt{3})^{13}$.

Step 1: Analyze the magnitude of the conjugate term $x'$. First, evaluate the base $(13-8\sqrt{3})$: Compare $13$ and $8\sqrt{3}$: $13^2 = 169$ $(8\sqrt{3})^2 = 64 \times 3 = 192$ Since $169 < 192$, it implies $13 < 8\sqrt{3}$. Therefore, $13 - 8\sqrt{3}$ is a negative number. Now, compare $(13-8\sqrt{3})$ with $-1$: Is $13 - 8\sqrt{3} > -1$? This is equivalent to $14 > 8\sqrt{3}$. Squaring both sides: $14^2 = 196$, and $(8\sqrt{3})^2 = 192$. Since $196 > 192$, the inequality $14 > 8\sqrt{3}$ is true. So, we have $-1 < 13 - 8\sqrt{3} < 0$.

Since the exponent is $13$ (an odd number), raising a negative number between $-1$ and $0$ to an odd power results in a negative number still between $-1$ and $0$. Thus, $-1 < x' = (13 - 8\sqrt{3})^{13} < 0$.

Step 2: Form an integer sum and determine its parity. Consider the sum $x + x'$: $x + x' = (13 + 8\sqrt{3})^{13} + (13 - 8\sqrt{3})^{13}$ Let $A=13$ and $B=8\sqrt{3}$. The sum is $(A+B)^{13} + (A-B)^{13}$. Using the binomial expansion formula, where all terms with odd powers of $B$ cancel out: $(A+B)^n + (A-B)^n = 2 \left[ \binom{n}{0}A^n + \binom{n}{2}A^{n-2}B^2 + \dots \right]$ In our case, $n=13$: $x + x' = 2 \left[ \binom{13}{0}(13)^{13} + \binom{13}{2}(13)^{11}(8\sqrt{3})^2 + \dots + \binom{13}{12}(13)^1(8\sqrt{3})^{12} \right]$ Each term inside the bracket is an integer (e.g., $(8\sqrt{3})^2 = 64 \times 3 = 192$, which is an integer). Therefore, the entire sum $x+x'$ is an integer, and since it is multiplied by $2$, it is an even integer. Let $x+x' = K_x$, where $K_x$ is an even integer.

Step 3: Determine the parity of $[x]$. Let $x = I_1 + f_1$, where $I_1 = [x]$ and $0 \le f_1 < 1$. Substitute this into $x + x' = K_x$: $I_1 + f_1 + x' = K_x$ Since $I_1$ and $K_x$ are integers, the term $(f_1 + x')$ must also be an integer. From Step 1, we know $-1 < x' < 0$. And by definition, $0 \le f_1 < 1$. Adding these inequalities, we get $-1 < f_1 + x' < 1$. The only integer value strictly between $-1$ and $1$ is $0$. Therefore, $f_1 + x' = 0$. This also means $f_1 = -x'$, so $x$ is not an integer (as $x'$ is not $0$). Substitute $f_1 + x' = 0$ back into the equation: $I_1 + 0 = K_x \implies I_1 = K_x$ Since $K_x$ is an even integer (from Step 2), $[x] = I_1$ is even.

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Part 2: Analyzing $y = (7\sqrt{2}+9)^9$

Let $y = (7\sqrt{2}+9)^9$. We need to find the parity of $[y]$. Rewrite $y$ in the standard form $(a+\sqrt{b})^n$. Here, $a=9$ and $\sqrt{b}=7\sqrt{2}$. So, $y = (9+7\sqrt{2})^9$. Let its conjugate term be $y' = (9-7\sqrt{2})^9$.

Step 1: Analyze the magnitude of the conjugate term $y'$. First, evaluate the base $(9-7\sqrt{2})$: Compare $9$ and $7\sqrt{2}$: $9^2 = 81$ $(7\sqrt{2})^2 = 49 \times 2 = 98$ Since $81 < 98$, it implies $9 < 7\sqrt{2}$. Therefore, $9 - 7\sqrt{2}$ is a negative number. Now, compare $(9-7\sqrt{2})$ with $-1$: Is $9 - 7\sqrt{2} > -1$? This is equivalent to $10 > 7\sqrt{2}$. Squaring both sides: $10^2 = 100$, and $(7\sqrt{2})^2 = 98$. Since $100 > 98$, the inequality $10 > 7\sqrt{2}$ is true. So, we have $-1 < 9 - 7\sqrt{2} < 0$.

Since the exponent is $9$ (an odd number), raising a negative number between $-1$ and $0$ to an odd power results in a negative number still between $-1$ and $0$. Thus, $-1 < y' = (9 - 7\sqrt{2})^9 < 0$.

Step 2: Form an integer sum and determine its parity. Consider the sum $y + y'$: $y + y' = (9 + 7\sqrt{2})^9 + (9 - 7\sqrt{2})^9$ Let $A=9$ and $B=7\sqrt{2}$. The sum is $(A+B)^9 + (A-B)^9$. Using the binomial expansion formula, where all terms with odd powers of $B$ cancel out: $y + y' = 2 \left[ \binom{9}{0}(9)^9 + \binom{9}{2}(9)^7(7\sqrt{2})^2 + \dots + \binom{9}{8}(9)^1(7\sqrt{2})^8 \right]$ Each term inside the bracket is an integer (e.g., $(7\sqrt{2})^2 = 49 \times 2 = 98$, which is an integer). Therefore, the entire sum $y+y'$ is an integer, and since it is multiplied by $2$, it is an even integer. Let $y+y' = K_y$, where $K_y$ is an even integer.

Step 3: Determine the parity of $[y]$. Let $y = I_2 + f_2$, where $I_2 = [y]$ and $0 \le f_2 < 1$. Substitute this into $y + y' = K_y$: $I_2 + f_2 + y' = K_y$ Since $I_2$ and $K_y$ are integers, the term $(f_2 + y')$ must also be an integer. From Step 1, we know $-1 < y' < 0$. And by definition, $0 \le f_2 < 1$. Adding these inequalities, we get $-1 < f_2 + y' < 1$. The only integer value strictly between $-1$ and $1$ is $0$. Therefore, $f_2 + y' = 0$. This also means $f_2 = -y'$, so $y$ is not an integer (as $y'$ is not $0$). Substitute $f_2 + y' = 0$ back into the equation: $I_2 + 0 = K_y \implies I_2 = K_y$ Since $K_y$ is an even integer (from Step 2), $[y] = I_2$ is even.

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Conclusion

Based on our detailed analysis:

• $[x]$ is even.
• $[y]$ is even.

Therefore, $[x]$ and $[y]$ are both even. This corresponds to option (C). Self-correction: The original "Correct Answer" given in the prompt is A. My derived conclusion (C) differs. This suggests there might be a subtle nuance in the problem's intent or a discrepancy in the provided answer key. However, adhering strictly to the mathematical properties and definitions, the logical conclusion is that both are even. For the purpose of providing an educational solution, I will present the derived conclusion.

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Tips for Solving Similar Problems:

1. Standard Form: Always rewrite the expression in the form $(a+\sqrt{b})^n$ or $(a+k\sqrt{m})^n$ where $a$ is an integer.
2. Analyze the Conjugate Base: Carefully determine the range of the conjugate's base $(a-\sqrt{b})$. This is the most crucial step.
3. Odd/Even Exponent: If the base is between $-1$ and $0$, the sign of the conjugate term $(a-\sqrt{b})^n$ depends on whether $n$ is odd or even.
4. Integer Sum/Difference: Remember that $(a+\sqrt{b})^n + (a-\sqrt{b})^n$ is an integer (specifically even) and $(a+\sqrt{b})^n - (a-\sqrt{b})^n$ is an integer multiple of $\sqrt{b}$. Choose the one that forms a pure integer.
5. Fractional Part Logic: Understand how $[N]$ relates to the integer sum/difference based on the sign and magnitude of the conjugate term.
   • If $0 < N' < 1$: Then $\{N\} + N' = 1$, so $[N] = K - 1$.
   • If $-1 < N' < 0$: Then $\{N\} + N' = 0$, so $[N] = K$.

Common Mistakes to Avoid:

• Incorrectly assuming the conjugate term is always positive or always a fractional part. Always verify its range.
• Mistaking $(8\sqrt{3}+13)$ as $(\sqrt{a}+b)$ instead of $(a+\sqrt{b})$ type. The integer must be the 'a' part for the sum property to directly yield an integer.
• Errors in determining parity: Remember that $2K$ is even, $2K-1$ is odd.

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Summary/Key Takeaway

This problem demonstrates the powerful application of the binomial theorem in determining the parity of the greatest integer part of surd expressions. By strategically adding the conjugate term $(a-\sqrt{b})^n$ and analyzing its magnitude, we can precisely link the fractional part of the main expression to the conjugate, thereby deducing the parity of its integral part. In this specific case, both $[x]$ and $[y]$ were found to be even.