Let A = {1, 2, 3, ..., 10} and R be a relation on A such that R = {(a, b) : a = 2b + 1}. Let (a1, a2

Q: Wait, the question says the set $A$ is $\{1, 2, \dots, 10\}$, but the options are all around 5 or 6. Did I misread something?

You have a sharp eye! This is a classic, well-known typo in this specific JEE problem. The set $A$ should indeed be $\{1, 2, \dots, 100\}$. If the set were only up to 10, the maximum $k$ would be much smaller. Always trust your intuition when the options don't align with the constraints—it's often a sign of a typo in the source material.

Q: Why do we set $a_{k+1} = 1$? Couldn't $a_{k+1}$ be something else?

That is a fantastic question. We want to maximize $k$, which means we want the longest possible chain. Our formula is $a_1 = 2^k a_{k+1} + 2^k - 1$. To keep $a_1$ within the limit of 100 while increasing $k$, we need to make the term $2^k a_{k+1}$ as small as possible. Since $a_{k+1}$ must be at least 1 (the smallest element in our set), setting $a_{k+1} = 1$ is the optimal strategy to allow $k$ to reach its maximum value.

Q: Why can't we just use a simple linear search to find the sequence?

You absolutely could, but that is the 'brute force' way! In JEE Advanced, we want to find the 'elegant' way. By deriving the general formula $a_1 = 2^{k+1} - 1$, we turn a tedious search into a simple inequality. This not only saves time but also protects you from missing a step in the chain. Mathematics is about finding the structure behind the chaos!

Question

Let $A = \{1, 2, 3, \dots, 10\}$ and $R$ be a relation on $A$ such that $R = \{(a, b) : a = 2b + 1\}$. Let $(a_1, a_2), (a_2, a_3), (a_3, a_4), \dots, (a_k, a_{k+1})$ be a sequence of elements of $R$ such that the second entry of an ordered pair is equal to the first entry of the next ordered pair. Then the largest integer $k$, for which such a sequence exists, is equal to :

Options:

(A) 6

(B) 7

(C) 5

(D) 8

Sigma Percentile · Accepted Answer

### Analyzing the Setup

My dear student, welcome to a journey through one of the most elegant problems in set theory and relations. Often, when we see a relation defined as $R = \{(a, b) : a = 2b + 1\}$, we might feel tempted to start listing pairs immediately.

But I want you to pause. I want you to see the structure. This is not just a collection of pairs; it is a machine. It is a generator.

When you have a sequence of pairs $(a_1, a_2), (a_2, a_3), \dots, (a_k, a_{k+1})$, you are not just looking at a list; you are looking at a chain reaction. Imagine each pair as a domino. When one falls, it triggers the next. The second element of one pair is the first element of the next. This linkage is the heartbeat of the problem.

### The Anatomy of the Recurrence

Let us translate the relation into the language of algebra. We are told that for any pair $(a_i, a_{i+1}) \in R$, the condition $a_i = 2a_{i+1} + 1$ must hold. This is our recurrence relation.

It tells us how to step backward from any element to the one that preceded it in the chain. If we know $a_2$, we know $a_1$. If we know $a_3$, we know $a_2$.

But what if we want to jump across the entire chain? Let us perform a substitution. We know $a_1 = 2a_2 + 1$ and $a_2 = 2a_3 + 1$.

If we substitute the second into the first, we get $a_1 = 2(2a_3 + 1) + 1$, which simplifies to $a_1 = 4a_3 + 3$. Do you see the pattern emerging? It is $2^2 a_3 + (2^2 - 1)$.

If we take one more step, $a_1 = 2^3 a_4 + (2^3 - 1)$. The beauty of mathematics lies in this predictability. We can generalize this for any $k$ steps:

$$a_1 = 2^k a_{k+1} + (2^k - 1)$$

This formula is your master key. It connects the very beginning of the chain to the very end, bypassing all the intermediate steps.

### The Constraint: The Boundary of Possibility

Now, we must respect the boundaries. The problem tells us that all elements must belong to the set $A = \{1, 2, \dots, 100\}$. This means $1 \le a_i \le 100$ for all $i$.

We want to find the largest integer $k$. To make $k$ as large as possible, we need to keep $a_1$ as small as possible, which in turn means we need to make $a_{k+1}$ as small as possible.

The smallest value any element in our set can take is $1$. So, let us set $a_{k+1} = 1$. Our master equation becomes:

$$a_1 = 2^k(1) + 2^k - 1 = 2^{k+1} - 1$$

Now, we apply our constraint: $a_1 \le 100$. This gives us the inequality:

$$2^{k+1} - 1 \le 100 \Rightarrow 2^{k+1} \le 101$$

This is the moment of truth. We are looking for the largest integer $k$ that satisfies this. We know that $2^6 = 64$, which is less than 101, but $2^7 = 128$, which exceeds our limit.

Thus, the maximum value for the exponent $k+1$ is $6$. Solving for $k$, we find **$k = 5$**.

### The Final Reflection

Think about what we just achieved. We didn't just solve a problem; we modeled a system. We took a complex, multi-step sequence and collapsed it into a single, powerful inequality.

When you look at the sequence $63, 31, 15, 7, 3, 1$, you are seeing the physical manifestation of our math. Each step is a perfect, logical progression.

Never fear the complexity of a problem, my friend. Instead, look for the pattern, embrace the recurrence, and let the algebra guide you to the truth. You have mastered the chain. Keep this clarity with you as you tackle the next challenge.

Select Answer:

Visualized Solution

Defining the Relation $R$

Structure of the Sequence

The Recurrence Relation

Expanding the Chain

Generalizing the Formula

Setting the Constraints

Solving the Inequality

Finding the Maximum $k$

The Way Forward

The Sigma Insight: Types of Relations

Analyzing the Setup

The Anatomy of the Recurrence

The Constraint

The Final Reflection

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