The number of divisors of $(2^6 \times 3^5 \times 5^3 \times 7^2)$, which are of the form $(3r+1)$, where $r$ is a non-negative integer, is

Every divisor of $2^6 \times 3^5 \times 5^3 \times 7^2$ has the form $2^i \times 3^j \times 5^k \times 7^l$

where $0 \leq i \leq 6, \quad 0 \leq j \leq 5, \quad 0 \leq k \leq 3, \quad 0 \leq l \leq 2$

For a divisor to be of the form $3r + 1$, it must leave remainder $1$ when divided by $3$.

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If $j \geq 1$, the divisor contains a factor of $3$, making it divisible by $3$, so the remainder would be $0$, not $1$.

Therefore $j = 0$, and the divisor reduces to $2^i \times 5^k \times 7^l$.

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Finding remainders of each base mod $3$:

$2 \equiv -1 \pmod{3} \implies 2^i \equiv (-1)^i \pmod{3}$

$5 \equiv -1 \pmod{3} \implies 5^k \equiv (-1)^k \pmod{3}$

$7 \equiv 1 \pmod{3} \implies 7^l \equiv 1 \pmod{3}$

So the condition $2^i \times 5^k \times 7^l \equiv 1 \pmod{3}$ becomes:

$(-1)^i \times (-1)^k \times 1 \equiv 1 \pmod{3}$

$(-1)^{i+k} \equiv 1 \pmod{3}$

This means $i + k$ must be even. The value of $l$ does not affect the remainder, so $l$ is free.

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$i + k$ is even when both are even or both are odd.

From $i \in \{0,1,2,3,4,5,6\}$: even values $= 4$, odd values $= 3$

From $k \in \{0,1,2,3\}$: even values $= 2$, odd values $= 2$

Number of valid $(i, k)$ pairs $= (4 \times 2) + (3 \times 2)$

$= 8 + 6$

$= 14$

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Since $l$ can be $0, 1,$ or $2$, there are $3$ choices for $l$.

Total divisors of the form $3r + 1$:

$= 14 \times 3$

$= 42$