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

Welcome, fellow traveler on this journey through the landscape of JEE Advanced mathematics. Today, we face a problem that, at first glance, seems designed to crush our spirits with its sheer scale.

We are asked to find the parity of the greatest integer parts of two monstrous expressions: $x=(8\sqrt{3}+13{)}^{13}$ and $y=(7\sqrt{2}+9{)}^9$. Calculating these values directly is a fool's errand.

But in mathematics, when brute force fails, elegance prevails. Let us peel back the layers of these expressions to reveal the simple, beautiful truth hidden underneath.

The Conjugate Strategy

Imagine you are standing before a locked door. The key is not in the expression itself, but in its shadow—its conjugate.

We define $x=I_1+f_1$, where $I_1$ is the integer part we seek, and $f_1$ is the fractional part, trapped between $0$ and $1$. Now, consider the conjugate $f_1^{\prime }=(8\sqrt{3}-13{)}^{13}$.

Why this specific form? Because of the binomial theorem. When we look at the difference $x+f_1^{\prime }$, we are essentially performing a dance of cancellation.

The binomial expansion creates a scenario where the terms with odd powers of $8\sqrt{3}$ vanish, leaving only the even powers. Since $(8\sqrt{3}{)}^2=192$, any even power of $8\sqrt{3}$ is an integer. Thus, $x+f_1^{\prime }$ must be an even integer, which we can call $2k$.

The Fractional Trap

Now, we must be rigorous. We have $x+f_1^{\prime }=2k$. Substituting our definition of $x$, we get $I_1+f_1+f_1^{\prime }=2k$.

Rearranging this, we find $f_1+f_1^{\prime }=2k-I_1$. The right side is clearly an integer.

But what about the left side? We know $0\le f_1<1$ and, crucially, $0<f_1^{\prime }<1$. Because $8\sqrt{3}=\sqrt{192}\approx 13.85$, the base $8\sqrt{3}-13$ is approximately $0.85$.

Raising this to the power of $13$ keeps it strictly between $0$ and $1$. Therefore, the sum $f_1+f_1^{\prime }$ must lie strictly between $0$ and $2$.

The only integer in this range is $1$. This forces $f_1+f_1^{\prime }=1$, and consequently, $I_1=2k-1$. Our integer part $I_1$ is odd.

The Final Victory

We apply this exact logic to $y=(7\sqrt{2}+9{)}^9$. We define its conjugate $f_2^{\prime }=(7\sqrt{2}-9{)}^9$.

Just as before, $7\sqrt{2}=\sqrt{98}\approx 9.89$, so $7\sqrt{2}-9\approx 0.89$, which is between $0$ and $1$. The binomial expansion of $y+f_2^{\prime }$ again yields an even integer, $2m$.

By the same logic of fractional sums, the fractional part of $y$ must satisfy $f_2+f_2^{\prime }=1$, and the integer part $I_2$ must be $2m-1$, which is odd.

We have arrived at the summit. We have two odd integers, $I_1$ and $I_2$. Their sum, $[x]+[y]=I_1+I_2$, must be even.

You see? The complexity was merely a facade. By understanding the structure of binomial expansions and the bounds of fractional parts, we have conquered the problem with grace and precision.