### Binomial Theorem Question 14

#### Question 14 - 30 January - Shift 2

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

(1) $[x]+[y]$ is even

(2) $[x]$ is odd but $[y]$ is even

(3) $[x]$ is even but $[y]$ is odd

(4) $[x]$ and $[y]$ are both odd

## Show Answer

#### Answer: (1)

#### Solution:

#### Formula: Binomial Theorem

$ \begin{gathered} x=(8 \sqrt{3}+13)={ }^{13} C_0 \cdot(8 \sqrt{3})^{13}+{ }^{13} C_1(8 \sqrt{3})^{12}(13)^{1}+\ldots \\ x^{\prime}=(8 \sqrt{3}-13)^{13}={ }^{13} C_0(8 \sqrt{3})^{13}-{ }^{13} C_1(8 \sqrt{3})^{12}(13)^{1}+\ldots \\ x-x^{\prime}=2[{ }^{13} C_1 \cdot(8 \sqrt{3})^{12}(13)^{1}+{ }^{13} C_3(8 \sqrt{3})^{10} \cdot(13)^{3} \ldots] \end{gathered} $

therefore, $x-x^{\prime}$ is even integer, hence $[x]$ is even

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} \ldots \ldots $

$y^{\prime}=(7 \sqrt{2}-9)^{9}={ }^{9} C_0(7 \sqrt{2})^{9}-{ }^{9} C_1(7 \sqrt{2})^{8}(9)^{1}+{ }^{9} C_2(7 \sqrt{2})^{7}(9)^{2} \ldots \ldots$

$y-y^{\prime}=2[{ }^{9} C_1(7 \sqrt{2})^{8}(9)^{1}+{ }^{9} C_3(7 \sqrt{2})^{6}(9)^{3}+\ldots]$

$y-y^{\prime}=$ Even integer, hence $[y]$ is even

Hence, $[x] + [y] $ is also even integer.