Let x = (8 sqrt{3} + 13)^{13} and y = (7 sqrt | vedclass

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(D) Let $x = (8 \sqrt{3} + 13)^{13}$ and $x' = (8 \sqrt{3} - 13)^{13}$. Since $0 < 8 \sqrt{3} - 13 < 1$,we have $0 < x' < 1$.$x + x' = (8 \sqrt{3} + 1

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$[x]$ and $[y]$ are both odd

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(D) Let $x = (8 \sqrt{3} + 13)^{13}$ and $x' = (8 \sqrt{3} - 13)^{13}$. Since $0 $x + x' = (8 \sqrt{3} + 13)^{13} + (8 \sqrt{3} - 13)^{13} = 2 \sum_{k=0, 2, 4, \dots}^{12} \binom{13}{k} (8 \sqrt{3})^{13-k} (13)^k$.This is an even integer. Since $x + x' = I$ (an even integer) and $0 Now,let $y = (7 \sqrt{2} + 9)^9$ and $y' = (7 \sqrt{2} - 9)^9$. Since $0 $y + y' = (7 \sqrt{2} + 9)^9 + (7 \sqrt{2} - 9)^9 = 2 \sum_{k=0, 2, 4, \dots}^{8} \binom{9}{k} (7 \sqrt{2})^{9-k} (9)^k$.This is an even integer. Since $y + y' = J$ (an even integer) and $0 Therefore,both $[x]$ and $[y]$ are odd.

Let $x = (8 \sqrt{3} + 13)^{13}$ and $y = (7 \sqrt{2} + 9)^9$. If $[t]$ denotes the greatest integer $\leq t$,then:

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