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 be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
- Let ABC be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side AB, five points $p_5, p_6, p_7, p_8, p_9$ on the side BC, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, ..., p_{13}$, is _________
- If $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is a solution of the system of equations $AX = B$, where $\text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$, then $|x + y + z|$ is equal to :
- Let $C_r$ denote the coefficient of $x^r$ in the binomial expansion of $(1 + x)^n$, $n \in \mathbb{N}, 0 \leq r \leq n$. If $P_n = C_0 - C_1 + \frac{2^2}{3} C_2 - \frac{2^3}{4} C_3 + \dots + \frac{(-2)^n}{n+1} C_n$, then the value of $\sum_{n=1}^{25} \frac{1}{P_{2n}}$ equals.
- The number of elements in the relation $R = \{(x, y) : 4x^2 + y^2 < 52, x, y \in \mathbb{Z}\}$ is
- Let $S = \{z \in \mathbb{C} : 4z^2 + \bar{z} = 0\}$. Then $\sum_{z \in S} |z|^2$ is equal to :
Let f and g be functions satisfying $f(x+y) = f(x)f(y), f(1) = 7$ and $g(x+y) = g(xy), g(1) = 1$, for all $x, y \in \mathbb{N}$. If $\sum_{x=1}^{n} \left(\frac{f(x)}{g(x)}\right) = 19607$, then n is equal to :
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.
- Let $S = \frac{1}{25!} + \frac{1}{3!23!} + \frac{1}{5!21!} + \dots$ up to 13 terms. If $13S = \frac{2^k}{n!}, k \in \mathbb{N}$, then $n + k$ is equal to
Important Questions from Algebra
- Let A be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
- Let ABC be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side AB, five points $p_5, p_6, p_7, p_8, p_9$ on the side BC, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, ..., p_{13}$, is _________
- If $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is a solution of the system of equations $AX = B$, where $\text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$, then $|x + y + z|$ is equal to :
- Let $C_r$ denote the coefficient of $x^r$ in the binomial expansion of $(1 + x)^n$, $n \in \mathbb{N}, 0 \leq r \leq n$. If $P_n = C_0 - C_1 + \frac{2^2}{3} C_2 - \frac{2^3}{4} C_3 + \dots + \frac{(-2)^n}{n+1} C_n$, then the value of $\sum_{n=1}^{25} \frac{1}{P_{2n}}$ equals.
