Let A=1,2,3, .,100 and R be a relation on A such that R=(a,b):a=2b+1.

Question

Let $A={1,2,3,\ldots .,100}$ and $R$ be a relation on $A$ such that $R={(a,b):a=2b+1}$. Let $({a}_{1}$, ${a}_{2}),({a}_{2},{a}_{3}),({a}_{3},{a}_{4}),\ldots .,({a}_{k},{a}_{k+1})$ be a sequence of $k$ 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 :

Accepted Answer

The relation $R$ is defined on the set $A={1,2,3,\ldots ,100}$ 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:

$({a}_{1},{a}_{2}),({a}_{2},{a}_{3}),\ldots ,({a}_{k},{a}_{k+1})$

Here, each ${a}_{i}$ satisfies the equation ${a}_{i}=2{a}_{i+1}+1$. Consequently, ${a}_{1}=2{a}_{2}+1$, making ${a}_{1}$ an odd number.

Let's examine the pattern:

${a}_{2}=2{a}_{3}+1$, implying ${a}_{1}=2(2{a}_{3}+1)+1=4{a}_{3}+3$.

${a}_{3}=2{a}_{4}+1$, leading to ${a}_{1}=4(2{a}_{4}+1)+3=8{a}_{4}+7$.

Continuing this pattern, we find:

${a}_{k}=2{a}_{k+1}+1\, \implies \, {a}_{1}={2}^{k}\cdot {a}_{k+1}+({2}^{k}-1)$

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

${a}_{k+1}=\frac{{a}_{1}+1-{2}^{k}}{{2}^{k}}$

Thus, ${2}^{k}|({a}_{1}+1)$. The task is to find the highest $k$ where ${2}^{k}$ divides any ${e}_{i}$ in ${2,\ldots ,101}$.

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:

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

Solutions · 1