The number 111...1 (91 times) is a/an

Let, $\begin{array}{l}\mathrm{S}=1+10+{10}^2+\dots +{10}^{90}\\ =\frac{1\cdot \left({10}^{91}-1\right)}{10-1}\\ =\frac{{\left({10}^{13}\right)}^7-1}{{10}^{13}-1}\times \frac{{10}^{13}-1}{10-1}\\ =\left[{\left({10}^{13}\right)}^6+{\left({10}^{13}\right)}^5+{\left({10}^{13}\right)}^4+\dots +1\right]\\ \times \left({10}^{12}+{10}^{11}+\dots +1\right)\end{array}$

∴ It is  the product  of two  integers  and hence  not prime.