Let N = 126 000 = 2⁴ · 3² · 5³ · 7. Count the positive divisors of N that satisfy each condition…

(a) The exponents have $5$, $3$, $4$ and $2$ choices: $5 \times 3 \times 4 \times 2 = 120$.

(b) For a square each exponent is even: $a \in \{0, 2, 4\}$, $b \in \{0, 2\}$, $c \in \{0, 2\}$, $d = 0$. That is $3 \times 2 \times 2 \times 1 = 12$.

(c) Write the divisor as $20m$. Then $m$ is a divisor of $N / 20 = 6300 = 2^2 \cdot 3^2 \cdot 5^2 \cdot 7$, which has $3 \times 3 \times 3 \times 2 = 54$ divisors.

(d) The sum is $(1 + 2 + 4 + 8 + 16)(1 + 3 + 9)(1 + 5 + 25 + 125)(1 + 7) = 31 \times 13 \times 156 \times 8 = 502944$.

(e) The divisor must contain neither $3$ nor $5$. So it divides $2^4 \cdot 7$: $5 \times 2 = 10$.