Question:
Let $\mathrm{n}$ 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 _____________
Solution:
$\mathrm{N}=2^{10} \times 5^{10} \times 11^{11} \times 13^{13}$
Now, power of 2 must be zero,
power of 5 can be anything,
power of 13 can be anything.
But, power of 11 should be even.
So, required number of divisors is
$1 \times 11 \times 14 \times 6=924$