Understanding the Relation and Sequence
The problem defines a set $A = \{1, 2, 3,...., 100\}$ and a relation $R = \{(a, b) : a=2b+1\}$. We are looking for the longest sequence of ordered pairs in $R$, denoted as $(a_1, a_2), (a_2, a_3), ..., (a_k, a_{k+1})$, where the second element of each pair matches the first element of the next pair. This implies a chain relationship: $a_i = 2a_{i+1}+1$ for $i = 1, 2, ..., k$. All elements $a_1, a_2, ..., a_{k+1}$ must belong to the set $A$. The relation $a = 2b+1$ means that $a$ must be an odd integer and $a > b$ for any pair $(a, b)$ in $R$. Since $a_i = 2a_{i+1}+1$, the sequence of elements $a_1, a_2, ..., a_{k+1}$ must be strictly decreasing.
Finding the Maximum Chain Length
To find the largest possible value of $k$ (the number of pairs), we need to construct the longest possible chain of elements $a_1, a_2, ..., a_{k+1}$ that are all within the set $A$. Since the sequence must decrease, we should start building the chain from the smallest possible valid value for $a_{k+1}$ and work backwards.
The smallest element in set $A$ is 1. Let's assume $a_{k+1} = 1$. We can then find the preceding elements using the relation $a_i = 2a_{i+1}+1$:
- If $a_{k+1} = 1$, then $a_k = 2(1) + 1 = 3$. ($3 \in A$)
- If $a_k = 3$, then $a_{k-1} = 2(3) + 1 = 7$. ($7 \in A$)
- If $a_{k-1} = 7$, then $a_{k-2} = 2(7) + 1 = 15$. ($15 \in A$)
- If $a_{k-2} = 15$, then $a_{k-3} = 2(15) + 1 = 31$. ($31 \in A$)
- If $a_{k-3} = 31$, then $a_{k-4} = 2(31) + 1 = 63$. ($63 \in A$)
- If $a_{k-4} = 63$, then $a_{k-5} = 2(63) + 1 = 127$.
Determining the Value of k
The value $a_{k-5} = 127$ is greater than 100, which means it is not in the set $A$. Therefore, the chain must terminate with the element 63. The longest sequence of elements within $A$ is $63, 31, 15, 7, 3, 1$. This sequence corresponds to the following ordered pairs in $R$:
- $(63, 31)$
- $(31, 15)$
- $(15, 7)$
- $(7, 3)$
- $(3, 1)$
This sequence contains 5 ordered pairs. The question defines $k$ as the number of these ordered pairs in the sequence $(a_1, a_2), (a_2, a_3), ..., (a_k, a_{k+1})$. Thus, the largest integer $k$ for which such a sequence exists is 5.
Similar Questions
Let A = {-3, -2, -1, 0, 1, 2, 3}. Let R be a relation on A defined by xRy if and only if $0\le x^2+2y\le 4$. Let $l$ be the number of elements in R and $m$ be the minimum number of elements required to be added in R to make it a reflexive relation. Then $l + m$ is equal to
Let $\alpha$ and $\beta$ be the roots of $x^2 + \sqrt{3}x-16=0$, and $\gamma$ and $\delta$ be the roots of $x^2 + 3x - 1 = 0$. If $P_n = \alpha^n + \beta^n$ and $Q_n = \gamma^n + \delta^n$, then $\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} + \frac{Q_{25}-Q_{23}}{Q_{24}}$ is equal to
Let the domain of the function $f (x) = \log_2 \log_4 \log_6(3 + 4x -x^2)$ be $(a, b)$.
If $\int_{b-a}^{b+a}[x^2] dx=p-\sqrt{q}-\sqrt{r}$, $p,q,r\in N$, $gcd(p,q,r) = 1$, where $[.]$ is the greatest integer function, then $p + q + r$ is equal to
Let A be a matrix of order $3 \times 3$ and $|A| = 5$. If $|2\text{adj} (3A \text{adj} (2A))| = 2^\alpha \cdot 3^\beta \cdot 5^\gamma$, $\alpha, \beta, \gamma \in N$, then $\alpha + \beta + \gamma$ is equal to
Let $a_1, a_2, a_3,....$ be a G.P. of increasing positive numbers. If $a_3a_5 = 729$ and $a_2 + a_4 = \frac{111}{4}$, then $24 (a_1 + a_2 + a_3)$ is equal to
All five letter words are made using all the letters A, B, C, D, E and arranged as in an English dictionary with serial numbers. Let the word at serial number $n$ be denoted by $W_n$. Let the probability $P(W_n)$ of choosing the word $W_n$ satisfy $P(W_n) = 2P(W_{n-1})$, $n > 1$.
If $P(CDBEA) = \frac{2^\alpha}{2^\beta-1}$, $\alpha, \beta\in N$, then $\alpha + \beta$ is equal to :
- The largest $n \in N$ such that $3^n$ divides $50!$ is :
- The number of sequences of ten terms, whose terms are either 0 or 1 or 2, that contain exactly five 1s and exactly three 2s, is equal to :
- Let A be the set of all functions $f: Z \to Z$ and R be a relation on A such that $R = \{(f, g) : f(0) = g(1) \text{ and } f(1) = g(0)\}$. Then R is :
Let $a \in \mathbf{R}$ and $A$ be a matrix of order $3 \times 3$ such that $\det (A) = -4$ and $A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix}$, where $I$ is the identity matrix of order $3 \times 3$. If $\det ((a+1)\text{adj}((a-1)A))$ is $2^m 3^n$, $m, n \in \{0, 1, 2, \dots, 20\}$, then $m+n$ is equal to :
Important Questions from Algebra
Let A = {-3, -2, -1, 0, 1, 2, 3}. Let R be a relation on A defined by xRy if and only if $0\le x^2+2y\le 4$. Let $l$ be the number of elements in R and $m$ be the minimum number of elements required to be added in R to make it a reflexive relation. Then $l + m$ is equal to
Let $\alpha$ and $\beta$ be the roots of $x^2 + \sqrt{3}x-16=0$, and $\gamma$ and $\delta$ be the roots of $x^2 + 3x - 1 = 0$. If $P_n = \alpha^n + \beta^n$ and $Q_n = \gamma^n + \delta^n$, then $\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} + \frac{Q_{25}-Q_{23}}{Q_{24}}$ is equal to
Let the domain of the function $f (x) = \log_2 \log_4 \log_6(3 + 4x -x^2)$ be $(a, b)$.
If $\int_{b-a}^{b+a}[x^2] dx=p-\sqrt{q}-\sqrt{r}$, $p,q,r\in N$, $gcd(p,q,r) = 1$, where $[.]$ is the greatest integer function, then $p + q + r$ is equal to
Let A be a matrix of order $3 \times 3$ and $|A| = 5$. If $|2\text{adj} (3A \text{adj} (2A))| = 2^\alpha \cdot 3^\beta \cdot 5^\gamma$, $\alpha, \beta, \gamma \in N$, then $\alpha + \beta + \gamma$ is equal to
Let $a_1, a_2, a_3,....$ be a G.P. of increasing positive numbers. If $a_3a_5 = 729$ and $a_2 + a_4 = \frac{111}{4}$, then $24 (a_1 + a_2 + a_3)$ is equal to
