The number of divisors of the form $4n + 1,$

$n\ge 0$ of the number is ${10}^{10}{11}^{11}{13}^{13}$ is

 

$2^{10}{5}^{10}{11}^{11}{13}^{13}$ has a divisor of the form 

$2^{\alpha }\cdot 5^{\beta }\cdot {11}^{ı}\cdot {13}^{\delta }$ where 

$\begin{array}{l}\alpha =0,1,2,\dots ,10,\beta =0,1,2,\dots ,10,\gamma =0,1,2,\dots ,11,\\ \delta =0,1,2,\dots ,13\end{array}$

It is of the form $4n+1$ if $\alpha =0;\beta =0,1,2,\dots ,10$

$\gamma =0,2,4,\dots ,10;\delta =0,1,2,\dots ,13$

The desired number is $11\times 6\times 14=924$