The number of divisors of $2^4 \times 3^5 \times 5^3 \times 7^2$ that are of the form $4r + 1$, where $r$ is a non-negative integer, is:
Answer and solution
Correct answer: A
Solution
A divisor of the form $4r + 1$ is odd, so it has no factor 2: $3^a \cdot 5^b \cdot 7^c$.
$5^b \equiv 1 \pmod 4$ always → 4 choices of $b$.
$3^a \equiv 3$ or $1$ and $7^c \equiv 3$ or $1$ (odd/even powers). Need $a + c$ even.
$a \in \{0..5\}$: 3 even, 3 odd; $c \in \{0,1,2\}$: 2 even, 1 odd → $3 \times 2 + 3 \times 1 = 9$
Total $= 9 \times 4 = 36$
$5^b \equiv 1 \pmod 4$ always → 4 choices of $b$.
$3^a \equiv 3$ or $1$ and $7^c \equiv 3$ or $1$ (odd/even powers). Need $a + c$ even.
$a \in \{0..5\}$: 3 even, 3 odd; $c \in \{0,1,2\}$: 2 even, 1 odd → $3 \times 2 + 3 \times 1 = 9$
Total $= 9 \times 4 = 36$