What is the maximum value of n such that 7 × 343 × 385 × 1000 × 2401 × 77777 is divisible by 35ⁿ ?

To find the maximum value of n such that the given product is divisible by 35ⁿ, we first need to prime factorize 35. 35 = 5 × 7

Next, we need to count the total number of factors of 5 and the total number of factors of 7 in the entire product. The maximum value of n will be the minimum of these two counts.

Let's break down each number in the product into its prime factors, focusing on 5 and 7:

1. 7 = 7¹ (contains one 7, zero 5s)
2. 343 = 7³ (contains three 7s, zero 5s)
3. 385 = 5 × 7 × 11 (contains one 5, one 7)
4. 1000 = 10³ = (2 × 5)³ = 2³ × 5³ (contains three 5s, zero 7s)
5. 2401 = 7⁴ (contains four 7s, zero 5s)
6. 77777 = 7 × 11111 = 7 × 41 × 271 (contains one 7, zero 5s)

Now, let's sum the counts for each prime factor:

Total number of factors of 7: 1 (from 7) + 3 (from 343) + 1 (from 385) + 0 (from 1000) + 4 (from 2401) + 1 (from 77777) = 10

Total number of factors of 5: 0 (from 7) + 0 (from 343) + 1 (from 385) + 3 (from 1000) + 0 (from 2401) + 0 (from 77777) = 4

The product can be expressed as (5⁴) × (7¹⁰) × (other prime factors). For the product to be divisible by 35ⁿ = (5 × 7)ⁿ = 5ⁿ × 7ⁿ, the value of n must be less than or equal to the total count of 5s and less than or equal to the total count of 7s. So, n ≤ 4 (from the count of 5s) And n ≤ 10 (from the count of 7s)

The maximum value of n is the minimum of (4, 10), which is 4.

Therefore, the maximum value of n is 4.

The final answer is B