Let $n$ be a non-negative integer. Then the number of divisors of the form " $4 n+1$ " of the number

Question

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

Professor PC Thomas Classes · Answer

(924) : Let $P=(10)^{10} \cdot(11)^{11} \cdot(13)^{13}$\ $=2^{10} \cdot 5^{10} \cdot 11^{11} \cdot 13^{13}$\ Now, any power of 2 can't be of the form $4 n+1$.\ Also, all powers of 5 and 13 are of the form $4 n+1$.\ Now, for 11, if we take even powers, then number would be of the form $4 n+1$.\ i.e., $\left(11^{0}, 11^{2}, 11^{4}, 11^{6}, 11^{8}, 11^{10}ight)$\ $	herefore \quad$ Total number of divisors $=(10+1)$\ $(13+1)(6)=11 	imes 14 	imes 6=924$\