Question

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

• 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 1

Answer: (B)

$x=(8\sqrt{3}+13)={}^{13}{C}_0\cdot (8\sqrt{3}{)}^{13}+{}^{13}{C}_1(8\sqrt{3}{)}^{12}(13{)}^1+\dots$
$x^{\prime }=(8\sqrt{3}-13{)}^{13}={}^{13}{C}_0(8\sqrt{3}{)}^{13}-{}^{13}{C}_1(8\sqrt{3}{)}^{12}(13{)}^1+\dots$
$x-x^{\prime }=2\left[{}^{13}{C}_1\cdot (8\sqrt{3}{)}^{12}(13{)}^1+{}^{13}{C}_3(8\sqrt{3}{)}^{10}\cdot (13{)}^3\dots \right]$
therefore, $x-x^{\prime }$ is even integer, hence $[x]$ is even
$\text{ Now, }y=(7\sqrt{2}+9{)}^9={}^9{C}_0(7\sqrt{2}{)}^9+{}^9{C}_1(7\sqrt{2}{)}^8(9{)}^1+{}^9{C}_2(7\sqrt{2}{)}^7(9{)}^2\dots \dots$
$y^{\prime }=(7\sqrt{2}-9{)}^9={}^9{C}_0(7\sqrt{2}{)}^9-{}^9{C}_1(7\sqrt{2}{)}^8(9{)}^1+{}^9{C}_2(7\sqrt{2}{)}^7(9{)}^2\dots \dots$
$y-y^{\prime }=2\left[{}^9{C}_1(7\sqrt{2}{)}^8(9{)}^1+{}^9{C}_3(7\sqrt{2}{)}^6(9{)}^3+\dots \right]$
$y-y^{\prime }=$ Even integer, hence $[y]$ is even