From the given sort of binary word, let length be 2k + 1. So no. of 1's = k + 1 and 0's=k. So the number, (say) n = $10^{2k} + 10^{2k - 2} + \cdots + 1$ $\implies n = \frac{10^{2k + 2} - 1}{99}$ $\implies n = \frac{(10^{k + 1} - 1)(10^{k + 1} + 1)}{99}$ $\implies n = \frac{(11\cdots11)(10^{k + 1} + 1)}{11}$ As n is prime if after dividing any one factor by 11, the quotient is greater than 1,then n becomes composite. So $111\cdots11(k$ times)$ = 11$ OR $10^{k + 1} + 1 = 11$ So in the first case we get k = 1 $\implies n = 101$ In the second case k = 0 $\implies n = 1$ As 1 is not prime, the only such $101\cdots01$ is $101$.