Let n be a non-negative integer. Then the num | vedclass

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(A) The given number is $N = (10)^{10} \cdot (11)^{11} \cdot (13)^{13} = 2^{10} \cdot 5^{10} \cdot 11^{11} \cdot 13^{13}$.$A$ divisor $d$ of $N$ is of

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$924$

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(A) The given number is $N = (10)^{10} \cdot (11)^{11} \cdot (13)^{13} = 2^{10} \cdot 5^{10} \cdot 11^{11} \cdot 13^{13}$.$A$ divisor $d$ of $N$ is of the form $2^a \cdot 5^b \cdot 11^c \cdot 13^d$,where $0 \le a \le 10, 0 \le b \le 10, 0 \le c \le 11, 0 \le d \le 13$.For $d$ to be of the form $4n+1$,$d$ must be odd,so $a=0$.Thus,$d = 5^b \cdot 11^c \cdot 13^d$.Modulo $4$,we have $5 \equiv 1 \pmod{4}$,$11 \equiv -1 \equiv 3 \pmod{4}$,and $13 \equiv 1 \pmod{4}$.So,$d \equiv 1^b \cdot (-1)^c \cdot 1^d \equiv (-1)^c \pmod{4}$.For $d \equiv 1 \pmod{4}$,we must have $(-1)^c = 1$,which means $c$ must be even.Possible values for $c \in \{0, 2, 4, 6, 8, 10\}$,which gives $6$ choices.Possible values for $b \in \{0, 1, 2, \dots, 10\}$,which gives $11$ choices.Possible values for $d \in \{0, 1, 2, \dots, 13\}$,which gives $14$ choices.The total number of such divisors is $11 	imes 6 	imes 14 = 924$.

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 $....$

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