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

The correct answer is 4.

To find the maximum value of n such that the product is divisible by 35ⁿ, we need to determine the prime factorization of each number in the expression and count the occurrences of the prime factors 5 and 7.

First, let's break down each number into its prime factors:

$7=7^1$

$343=7^3$

$385=5\times 7\times 11$

$1000=2^3\times 5^3$

$2401=7^4$

$77777=7\times 11111$

Now, let's sum the powers of the relevant prime factors (5 and 7) across the entire product:

1. Total power of 5:

The factor 5 appears in 385 (power of 1) and 1000 (power of 3).

$\text{Total count of 5}=1+3=4$

2. Total power of 7:

The factor 7 appears in 7 (power of 1), 343 (power of 3), 385 (power of 1), 2401 (power of 4), and 77777 (power of 1).

$\text{Total count of 7}=1+3+1+4+1=10$

Since $35=5\times 7$, a power of 35 can be written as:

${35}^n=5^n\times 7^n$

For the expression to be divisible by 35ⁿ, n must be less than or equal to the total available count of both 5 and 7. Therefore, n is limited by the smaller count between the factors 5 and 7:

$n=\text{min}(4,10)=4$

Thus, the maximum value of n is 4.