Question

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

Hint:

To find the number of divisors of a certain form (e.g., $4k+1$), analyze the prime factors of the number modulo 4. Primes of the form $4m+1$ can have any power, while primes of the form $4m+3$ must have an even power for the divisor to be of the form $4k+1$.

Answer 1

Correct Answer: 924

Key Steps:
\[ N=2^{10}5^{10}11^{11}13^{13} \] For divisors of form $4k+1$: - power of 2 must be zero - power of 11 must be even \[ \text{Count}=1\times11\times6\times14=924 \] \[ \boxed{924} \]