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

Correct Option (B)

To determine the maximum value of n such that the given product is divisible by 35ⁿ, we first express the divisor 35 in its prime factors. Since 35 = 5 × 7, it follows that 35ⁿ = 5ⁿ × 7ⁿ. For the product to be divisible by 35ⁿ, it must contain at least n factors of 5 and n factors of 7.

We proceed by performing the prime factorization for each term in the given product:

• 7 = 7¹
• 343 = 7³
• 385 = 5¹ × 7¹ × 11¹
• 1000 = 2³ × 5³
• 2401 = 7⁴
• 77777 = 7¹ × 41¹ × 271¹

Next, we calculate the total count of each prime factor (5 and 7) across all terms in the product:

• Total power of 5: The factors of 5 appear in 385 (5¹) and 1000 (5³). The sum of their exponents is 1 + 3 = 4.
• Total power of 7: The factors of 7 appear in 7 (7¹), 343 (7³), 385 (7¹), 2401 (7⁴), and 77777 (7¹). The sum of their exponents is 1 + 3 + 1 + 4 + 1 = 10.

For the product to be divisible by 5ⁿ × 7ⁿ, 'n' must be less than or equal to the minimum of the total powers of 5 and 7 available in the product. Therefore, n = min(4, 10) = 4.

The maximum value of n is 4.

Incorrect Options:

The maximum value of n is constrained by the prime factor with the lowest total power, which is 5 with a total power of 4. Consequently, any value of n greater than 4 (such as 5 or 7, corresponding to options C and D) would imply that the product must contain more factors of 5 than are actually present (only 4 factors of 5 are available). Option A (3) is incorrect because while the product is divisible by 35³, the question specifically asks for the maximum possible value for n, which is 4.