Let n be a non-negative integer. Then the number of divisors of the form “4n + 1” of the number ${\left(10\right)}^{10}.{\left(11\right)}^{11}.{\left(13\right)}^{13}$ is equal to

$$\begin{gathered} \mathrm{N}={10}^{10} {11}^{11} {13}^{13} \\ =2^{10}\times 5^{10}\times {11}^{11}\times {13}^{13} \\ \mathrm{Divisors} \mathrm{of} \mathrm{the} \mathrm{form} 4\mathrm{k}+1 \\ =\left(5^0+5^1+5^2+\dots \dots +5^{10}\right)({11}^0+{11}^2+{11}^4+\dots \dots +{11}^{10})\left({13}^0+{13}^1+{13}^2+...{13}^{13}\right) \\ =11\times 6\times 14 \\ =924 \end{gathered}$$