We need the number of divisors of the form $4n+1$ of $N=(10)^{10}(11)^{11}(13)^{13}.$

1. Prime factorization of $N$

First factor each base: $10=2\cdot 5,\qquad 11=11,\qquad 13=13.$ So, $N=10^{10}\cdot 11^{11}\cdot 13^{13}=2^{10}5^{10}11^{11}13^{13}.$

Any divisor $d$ of $N$ has the form $d=2^a5^b11^c13^e$ where $0\le a,b\le 10,\quad 0\le c\le 11,\quad 0\le e\le 13.$

We want those divisors for which $d\equiv 1\pmod 4.$

2. Residues modulo $4$

Now reduce each prime modulo $4$: $2\equiv 2 \pmod 4,\quad 5\equiv 1 \pmod 4,\quad 11\equiv 3 \pmod 4,\quad 13\equiv 1 \pmod 4.$

Thus,

• $5^b\equiv 1 \pmod 4$ for all $b$,
• $13^e\equiv 1 \pmod 4$ for all $e$,
• $11^c\equiv 3^c \pmod 4$.

Also, divisors of the form $4n+1$ must be odd. Hence $a=0$; otherwise $d$ is even and cannot be $1 \pmod 4$.

So we only need to count odd divisors $d=5^b11^c13^e.$

For these, $d\equiv 11^c\equiv 3^c \pmod 4.$ Now,