Prime Factorization of 105
To solve this problem, we need to first find the prime factorization of 105:

105 = 3 x 5 x 7

Number of Divisors
The number of divisors of a number can be found by adding 1 to each exponent in its prime factorization and then multiplying the results. In this case:

Number of divisors of 105 = (1 + 1)(1 + 1)(1 + 1) = 2 x 2 x 2 = 8

Counting Divisors with at Least One Zero
To count the number of divisors with at least one zero at the end, we need to consider the factors of 105 that contain at least one 2 and one 5.

Powers of 2
The powers of 2 that divide 105 are 2^0 and 2^1.

2^0 = 1 does not contribute a factor of 10 to any divisor, so we can ignore it.

2^1 = 2 contributes a factor of 10 to half of the divisors of 105.

Powers of 5
The powers of 5 that divide 105 are 5^0 = 1 and 5^1 = 5.

5^0 = 1 does not contribute a factor of 10 to any divisor, so we can ignore it.

5^1 = 5 contributes a factor of 10 to one-fifth of the divisors of 105.

Counting Divisors
To count the number of divisors with at least one zero at the end, we need to multiply the number of divisors of 105 by the fractions of divisors that have a factor of 10:

Number of divisors with a factor of 2 and 5 = 8 x (1/2) x (1/5) = 8/10 = 4/5

Number of divisors with at least one zero at the end = 4/5 x 8 = 32/5 ≈ 6.4

Since we are looking for a whole number of divisors, we need to round up to the nearest integer:

Number of divisors with at least one zero at the end ≈ 7

Therefore, the correct answer is option 'D' (25).