Let n be a non-negative integer.

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:

Answer: 924

Accepted Answer

$N = 10^{10}\cdot11^{11}\cdot13^{13} = 2^{10}\cdot5^{10}\cdot11^{11}\cdot13^{13}$.

For a divisor to be $\equiv1\pmod4$: the power of $2$ must be $0$ (since any factor of $2$ makes the divisor even, and we need odd form $4n+1$). Power of $5$: any (contributes $5\equiv1\pmod4$, so power doesn't affect residue class as long as we track total). Power of $13$: $13\equiv1\pmod4$, so any power is fine. Power of $11$: $11\equiv3\pmod4$, so we need the power of 11 to be even.

Number of choices: power of 2 fixed at 0 ($1$ way); power of 5: $0$ to $10$ ($11$ ways); power of 13: $0$ to $13$ ($14$ ways); power of 11 even: $0,2,4,6,8,10$ ($6$ ways).

$$\text{Total} = 1\times11\times14\times6 = 924$$

Solutions · 1