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

Question

Let $A=\{1,2,3, \ldots ., 100\}$ and $R$ be a relation on $A$ such that $R=\{(a, b): a=2 b+1\}$. Let $\left(a_1\right.$, $\left.a_2\right),\left(a_2, a_3\right),\left(a_3, a_4\right), \ldots .,\left(a_k, a_{k+1}\right)$ 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 = 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 \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 \mid (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) $

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