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

The divisors of the given number will have the form $$2^a*3^b*5^c*7^d$$ with $$0\le a\le6,\ 0\le b\le5,\ 0\le c\le3,\ 0\le d\le2$$

Because the divisors should be in the form 3r+1, it cannot be divisible by 3, so (b=0). 

Reduce modulo 3: $$2  \text{mod}  3 = 2, 5  \text{mod}  3 = 2, 7  \text{mod}  3 = 1$$ 

Hence,  $$2^a5^c7^d\equiv 2^{a+c}\cdot1^d \equiv 2^{a+c}\pmod3$$

2^k will be in the form 3r+1 only when K is even. So, we need a+c to be even

$$a\in{0,\dots,6}$$ has 4 even, 3 odd values 

$$c\in{0,\dots,3}$$ has 2 even, 2 odd

Number of (a,c) with (a+c) even is $$4\cdot2 + 3\cdot2 = 8+6=14$$

For each such pair, there are 3 choices for d = 0,1,2. Thus, total divisors in the form (3r+1) equals $$14\times3=42$$.