JEE Main 2025 — Sets And Relations Question with Solution

The relation $R$ is defined on the set $A=\{1,2,3,\dots ,100\}$ such that $R=\{(a,b):a=2b+1\}$. We need to find the largest integer $k$ for which there exists a sequence of $k$ ordered pairs from $R$ where the second element of each pair is the first element of the next pair.

The sequence in terms of $k$ is:

$(a_1,a_2),(a_2,a_3),\dots ,(a_k,a_{k+1})$

Here, each $a_i$ satisfies the equation $a_i=2a_{i+1}+1$. Consequently, $a_1=2a_2+1$, making $a_1$ an odd number.

Let's examine the pattern:

$a_2=2a_3+1$, implying $a_1=2(2a_3+1)+1=4a_3+3$.

$a_3=2a_4+1$, leading to $a_1=4(2a_4+1)+3=8a_4+7$.

Continuing this pattern, we find:

$a_k=2a_{k+1}+1⟹a_1=2^k\cdot a_{k+1}+(2^k-1)$

where $a_{k+1}$ needs to be in set $A$. This implies:

$a_{k+1}=\frac{a_1+1-2^k}{2^k}$

Thus, $2^k\mid (a_1+1)$. The task is to find the highest $k$ where $2^k$ divides any $e_i$ in $\{2,\dots ,101\}$.

The largest power of 2 that divides an element within this range determines $k$.

After computation, we find that $k$ can be a maximum of 6 because $2^6=64$ divides $95+1=96$, but $2^7=128$ does not divide any $e_i$ for $e_i\in A$. Therefore, the maximum $k$ is 6.

The sequence corresponding to this maximum $k$ is:

$(95,47),(47,23),(23,11),(11,5),(5,2)$