Step 1: Understand the form $(3r+1)$
Divisors of the form $3r+1$ are those which leave a remainder of 1 when divided by 3. If a divisor contains any power of 3 (i.e., $3^1, 3^2...$), it will be divisible by 3 ($3r$). Thus, we must take $3^0$ from the prime factor $3^5$.
Step 2: Analyze remaining prime factors $\pmod 3$
We are looking at divisors of $N' = 2^6 \cdot 5^3 \cdot 7^2$.
$2 \equiv -1 \pmod 3$
$5 \equiv -1 \pmod 3$
$7 \equiv 1 \pmod 3$
Step 3: Combinations for Remainder 1
Any power of 7 ($7^0, 7^1, 7^2$) is always $\equiv 1 \pmod 3$. So they don't change the remainder.
The product of $2^a \cdot 5^b$ must be $\equiv 1 \pmod 3$.
$(-1)^a \cdot (-1)^b = (-1)^{a+b}$. For this to be $1$, $a+b$ must be even.
Possible values: $a \in \{0,1..6\}$ (7 values), $b \in \{0,1,2,3\}$ (4 values).
Total pairs $(a,b) = 7 \times 4 = 28$. In any sequence of integers, exactly half the sums are even and half are odd if the total count is even. Here $28/2 = 14$ pairs.
Total divisors = $14 \text{ (from 2 and 5)} \times 3 \text{ (from 7)} = \mathbf{42}$.
