The number of divisors of (2⁶ × 3⁵ × 5³ × 7² ) , which are of the form (3r + 1) , where r is a non-negative integer, is

The correct answer is 42.

We are given the number $N=2^6\times 3^5\times 5^3\times 7^2$, and we need to find the number of its divisors that are of the form $3r+1$, where $r$ is a non-negative integer. A divisor of this form leaves a remainder of 1 when divided by 3, which means it is congruent to 1 modulo 3.

Any divisor $d$ of $N$ can be written in the form:

$d=2^a\times 3^b\times 5^c\times 7^k$

where the powers must satisfy the conditions $0\le a\le 6$, $0\le b\le 5$, $0\le c\le 3$, and $0\le k\le 2$.

Let's analyze the remainder of each prime factor when divided by 3:

1. For the factor 3: If $b>0$, the divisor $d$ will be a multiple of 3. A multiple of 3 can never leave a remainder of 1 when divided by 3. Therefore, to have $d\equiv 1(\mathrm{mod}3)<!--\; </mo-->$, we must strictly have $b=0$. This leaves us with 1 choice for the exponent $b$.

2. For the factor 2: Notice that $2\equiv -1(\mathrm{mod}3)<!--\; </mo-->$. Thus, $2^a\equiv {(-1)}^a(\mathrm{mod}3)<!--\; </mo-->$.

3. For the factor 5: Notice that $5\equiv 2\equiv -1(\mathrm{mod}3)<!--\; </mo-->$. Thus, $5^c\equiv {(-1)}^c(\mathrm{mod}3)<!--\; </mo-->$.

4. For the factor 7: Notice that $7\equiv 1(\mathrm{mod}3)<!--\; </mo-->$. Thus, $7^k\equiv 1^k\equiv 1(\mathrm{mod}3)<!--\; </mo-->$ for any power of $k$.

Putting this all together, the divisor $d$ modulo 3 is given by:

$d\equiv {(-1)}^a\times 1\times {(-1)}^c\times 1\equiv {(-1)}^{a+c}(\mathrm{mod}3)<!--\; </mo-->$

For the divisor $d$ to be of the form $3r+1$, we need $d\equiv 1(\mathrm{mod}3)<!--\; </mo-->$. This requires ${(-1)}^{a+c}=1$. This means that the sum $(a+c)<!--\; </mo-->$ must be an even integer. For the sum of two numbers to be even, they must both have the same parity (either both are even or both are odd).

Case 1: Both $a$ and $c$ are even.
The possible values for $a$ are from the set {0, 1, 2, 3, 4, 5, 6}. The even values are {0, 2, 4, 6}, which gives 4 choices.
The possible values for $c$ are from the set {0, 1, 2, 3}. The even values are {0, 2}, which gives 2 choices.
Total combinations for Case 1 = $4\times 2=8$.

Case 2: Both $a$ and $c$ are odd.
The odd values for $a$ are {1, 3, 5}, which gives 3 choices.
The odd values for $c$ are {1, 3}, which gives 2 choices.
Total combinations for Case 2 = $3\times 2=6$.

Summing the possibilities from both cases, there are $8+6=14$ valid combinations for the pair $(a,c)<!--\; </mo-->$.

Finally, we consider the exponent $k$ for the prime 7. Since 7 is congruent to 1 modulo 3, the exponent $k$ can be any valid value {0, 1, 2}. This gives 3 independent choices.

To find the total number of divisors of the form $3r+1$, we multiply the number of choices for the pair $(a,c)<!--\; </mo-->$ by the number of choices for $k$ (and remembering that $b$ has exactly 1 choice, 0):

$\text{Total Divisors}=14\times 3=42$

Therefore, there are exactly 42 divisors satisfying the condition.