JEE Mathematics 2023 Question with Solution

Given: $x = \left( 8\sqrt{3}+13 \right)^{13}$ and $y = \left( 7\sqrt{2}+9 \right)^9$.

Find: The correct statement about the parity of $[x]$ and $[y]$.

Take the conjugate-type expressions

$$x' = \left( 13 - 8\sqrt{3} \right)^{13}, \qquad y' = \left( 9 - 7\sqrt{2} \right)^9$$

Since $0 < 13 - 8\sqrt{3} < 1$ and $0 < 9 - 7\sqrt{2} < 1$, we get

$$0 < x' < 1, \qquad 0 < y' < 1$$

Now

$$x + x' = \left( 13 + 8\sqrt{3} \right)^{13} + \left( 13 - 8\sqrt{3} \right)^{13}$$

In the binomial expansion, all irrational terms cancel, so $x + x'$ is an integer. Also, every surviving term contains a factor $13$, hence $x + x'$ is odd.

Because $0 < x' < 1$, we have

$$x = (x + x') - x'$$

so $[x] = x + x' - 1$. Therefore, $[x]$ is even.

Similarly,

$$y + y' = \left( 9 + 7\sqrt{2} \right)^9 + \left( 9 - 7\sqrt{2} \right)^9$$

Again, irrational terms cancel, so this is an integer. Every surviving term has a factor $9$, hence $y + y'$ is odd.

Since $0 < y' < 1$, we get

$$[y] = y + y' - 1$$

Thus $[y]$ is even.

Therefore, both $[x]$ and $[y]$ are even, so $[x] + [y]$ is even.

The solution states the working leads to this conclusion, but its option label announcement says B, which conflicts with the actual conclusion. Hence the defensible correct option from the working is A.

Given: Expressions of the form $\left(a+b\sqrt{n}\right)^m$.

Find: The parity of their greatest integer parts.

Use the fact that if

$$0 < a-b\sqrt{n} < 1$$

then

$$\left(a+b\sqrt{n}\right)^m + \left(a-b\sqrt{n}\right)^m$$

is an integer, while the second term lies between $0$ and $1$. So the floor of the first term is exactly one less than that integer.

For both given expressions, this associated integer is odd, so the floor becomes odd minus $1$, which is even.

Hence $[x]$ and $[y]$ are both even, and the correct option is A.