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

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Permutations and Combinations

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)^{10} \cdot (13)^{13}

is equal to ____

Answer: 924

The number is

1 0^{10} \cdot 1 1^{10} \cdot 1 3^{13} = 2^{10} \cdot 5^{10} \cdot 1 1^{10} \cdot 1 3^{13}

, so every divisor has the form

2^{p} 5^{q} 1 1^{r} 1 3^{s}

with

0 \leq p \leq 10

0 \leq q \leq 10

0 \leq r \leq 10

and

0 \leq s \leq 13

. A number of the form

4 n + 1

is odd, so

p = 0

.
Modulo

4

5 \equiv 1

13 \equiv 1

and

11 \equiv - 1

, hence

5^{q} 1 1^{r} 1 3^{s} \equiv (- 1)^{r} (mod 4)

which equals

1

precisely when

r

is even. So

r \in {0, 2, 4, 6, 8, 10}

, giving

6

choices, while

q

and

s

are unrestricted with

11

and

14

choices.
The number of such divisors is

6 \times 11 \times 14 = 924

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Integer answer