Let there be \(100\) brilliant members labelled \(a_1, a_2, a_3, a_4, a_5,\ldots, a_{99}, a_{100}\) In this group of brilliant members, \(a_2\) follows \(a_1\), \(a_3\) follows \(a_2\), \(\ldots\) \(a_n\) follows \(a_{n-1}\), till \(a_{100}\) follows \(a_{99}\). Other than this group of members, these members have other followers, denoted by \(F\) who themselves have no followers. Now, \(a_n\) has \(n\) followers. Thus, for \(a_1\), the only follower is \(a_2\), while for \(a_2\), he has two followers: \(a_3\) and \(1 F\). This pattern continues till \(a_{100}\).

Now, say \(a_1\) makes a post, the total number of possible reshares which can be done by \(a_2, a_3, a_4,\ldots a_{99}, a_{100}\) and the \(F\)s is given by \(x\). All of the \(F\)s will definitely reshare a post made or reshared by the person(s) they follow, and all \(a_{2n+1}\), with \(n\in\mathbb{N}\), will do the same thing. However, for all \(a_{2n}\), with \(n\in\mathbb{N}\), the probability that they will reshare something posted by someone they follow is \(\frac12\). The probability that the number of reshares is \(\geq\frac{x}{2}\) is given by \(\frac{1}{a}\). Find the last digit of \(a\)

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