Question:

Step 1: Analyze $x$ and its conjugate term. Let the given number be $x=(8\sqrt{3}+13)^{13}$. To find its greatest integer part, we consider its conjugate. Let $x' = (13-8\sqrt{3})^{13}$. We first determine the range of $x'$. We compare $13$ and $8\sqrt{3}$: $13^2 = 169$ $(8\sqrt{3})^2 = 64 \times 3 = 192$ Since $192 > 169$, we have $8\sqrt{3} > 13$. Therefore, $8\sqrt{3}-13 > 0$. Approximating $8\sqrt{3} \approx 8 \times 1.732 = 13.856$. So, $0 < 8\sqrt{3}-13 < 1$. Let $k = 8\sqrt{3}-13$. Thus, $0 < k < 1$. Now, $13-8\sqrt{3} = -(8\sqrt{3}-13) = -k$. Since the exponent is an odd number ($13$), we have $x' = (-k)^{13} = -k^{13}$. As $0 < k < 1$, it follows that $0 < k^{13} < 1$. Therefore, $-1 < -k^{13} < 0$, which means $-1 < x' < 0$. Step 2: Evaluate the sum $x+x'$. Consider the sum $x+x'$: $$x+x' = (13+8\sqrt{3})^{13} + (13-8\sqrt{3})^{13}$$ Using the binomial expansion $(a+b)^n + (a-b)^n = 2\left[ \binom{n}{0}a^n + \binom{n}{2}a^{n-2}b^2 + \dots \right]$. Here, $a=13$, $b=8\sqrt{3}$, and $n=13$. All terms in the expansion will be integers, as $(8\sqrt{3})^2 = 192$ is an integer. $$x+x' = 2 \left[ \binom{13}{0}(13)^{13} + \binom{13}{2}(13)^{11}(8\sqrt{3})^2 + \dots + \binom{13}{12}(13)^1(8\sqrt{3})^{12} \right]$$ Since this sum is twice a sum of integers, $x+x'$ is an even integer. Let $K_1 = x+x'$. Step 3: Determine the parity of $[x]$. Let $[x]$ be the greatest integer less than or equal to $x$, and $\{x\}$ be the fractional part of $x$, such that $x = [x] + \{x\}$, where $0 \leq \{x\} < 1$. From Step 1, we know $-1 < x' < 0$. The sum $K_1 = x+x'$ can be written as $K_1 = ([x] + \{x\}) + x'$. Since $K_1$ is an integer, $\{x\} + x'$ must also be an integer. Given $0 \leq \{x\} < 1$ and $-1 < x' < 0$, we have $-1 < \{x\} + x' < 1$. For $\{x\} + x'$ to be an integer within this range, it must be $0$. So, $\{x\} + x' = 0$, which implies $\{x\} = -x'$. From Step 1, $x' = -k^{13}$, so $\{x\} = k^{13}$. Since $0 < k < 1$, we have $0 < k^{13} < 1$, which means $\{x\} \neq 0$. Therefore, $x$ is not an integer. Substituting $\{x\} + x' = 0$ into the equation for $K_1$, we get $K_1 = [x] + 0 = [x]$. Since $K_1$ is an even integer (from Step 2), $[x]$ must be an even integer. Step 4: Analyze $y$ and its conjugate term. Let the given number be $y=(7\sqrt{2}+9)^9$. Let's write it as $(9+7\sqrt{2})^9$. To find its greatest integer part, we consider its conjugate. Let $y' = (9-7\sqrt{2})^9$. We first determine the range of $y'$. We compare $9$ and $7\sqrt{2}$: $9^2 = 81$ $(7\sqrt{2})^2 = 49 \times 2 = 98$ Since $98 > 81$, we have $7\sqrt{2} > 9$. Therefore, $7\sqrt{2}-9 > 0$. Approximating $7\sqrt{2} \approx 7 \times 1.414 = 9.898$. So, $0 < 7\sqrt{2}-9 < 1$. Let $k' = 7\sqrt{2}-9$. Thus, $0 < k' < 1$. Now, $9-7\sqrt{2} = -(7\sqrt{2}-9) = -k'$. Since the exponent is an odd number ($9$), we have $y' = (-k')^9 = -k'^9$. As $0 < k' < 1$, it follows that $0 < k'^9 < 1$. Therefore, $-1 < -k'^9 < 0$, which means $-1 < y' < 0$. Step 5: Evaluate the sum $y+y'$. Consider the sum $y+y'$: $$y+y' = (9+7\sqrt{2})^9 + (9-7\sqrt{2})^9$$ Using the binomial expansion $(a+b)^n + (a-b)^n = 2\left[ \binom{n}{0}a^n + \binom{n}{2}a^{n-2}b^2 + \dots \right]$. Here, $a=9$, $b=7\sqrt{2}$, and $n=9$. All terms in the expansion will be integers, as $(7\sqrt{2})^2 = 98$ is an integer. $$y+y' = 2 \left[ \binom{9}{0}9^9 + \binom{9}{2}9^7(7\sqrt{2})^2 + \dots + \binom{9}{8}9^1(7\sqrt{2})^8 \right]$$ Since this sum is twice a sum of integers, $y+y'$ is an even integer. Let $K_2 = y+y'$. Step 6: Determine the parity of $[y]$. Let $[y]$ be the greatest integer less than or equal to $y$, and $\{y\}$ be the fractional part of $y$, such that $y = [y] + \{y\}$, where $0 \leq \{y\} < 1$. From Step 4, we know $-1 < y' < 0$. The sum $K_2 = y+y'$ can be written as $K_2 = ([y] + \{y\}) + y'$. Since $K_2$ is an integer, $\{y\} + y'$ must also be an integer. Given $0 \leq \{y\} < 1$ and $-1 < y' < 0$, we have $-1 < \{y\} + y' < 1$. For $\{y\} + y'$ to be an integer within this range, it must be $0$. So, $\{y\} + y' = 0$, which implies $\{y\} = -y'$. From Step 4, $y' = -k'^9$, so $\{y\} = k'^9$. Since $0 < k' < 1$, we have $0 < k'^9 < 1$, which means $\{y\} \neq 0$. Therefore, $y$ is not an integer. Substituting $\{y\} + y' = 0$ into the equation for $K_2$, we get $K_2 = [y] + 0 = [y]$. Since $K_2$ is an even integer (from Step 5), $[y]$ must be an even integer. Step 7: Conclude the parity of $[x]+[y]$. From Step 3, we determined that $[x]$ is an even integer. From Step 6, we determined that $[y]$ is an even integer. The sum of two even integers is an even integer. Therefore, $[x]+[y]$ is an even integer. The final answer is $\boxed{[x]+[y] \text{ is even}}$.