Understanding Divisibility and Prime Factorization

The question asks for the maximum integer value of \(n\) such that the product \(7 \times 343 \times 385 \times 1000 \times 2401 \times 77777\) is completely divisible by \(35^n\). To solve this divisibility problem, we need to use the concept of prime factorization. The divisor is \(35^n\). We know that \(35 = 5 \times 7\). Therefore, \(35^n = (5 \times 7)^n = 5^n \times 7^n\). For a number to be divisible by \(5^n \times 7^n\), its prime factorization must contain at least \(n\) factors of 5 and at least \(n\) factors of 7.

Let's find the prime factorization of each number in the given product:

• \(7 = 7^1\)
• \(343 = 7 \times 49 = 7 \times 7^2 = 7^3\)
• \(385\). This number ends in 5, so it's divisible by 5. \(385 = 5 \times 77 = 5 \times 7 \times 11\)
• \(1000 = 10^3 = (2 \times 5)^3 = 2^3 \times 5^3\)
• \(2401\). We can test divisibility by small prime numbers. Let's try 7: \(2401 \div 7 = 343\). We already found \(343 = 7^3\). So, \(2401 = 7 \times 343 = 7 \times 7^3 = 7^4\)
• \(77777\). This number is divisible by 7: \(77777 = 7 \times 11111\). Now we need to factorize 11111. Let's test small prime numbers: not divisible by 2, 3 (sum of digits is 5), 5. \(11111 \div 7\) is not an integer. \(11111 \div 11\) is not an integer. Let's try 41: \(11111 \div 41 = 271\). 271 is a prime number. So, \(77777 = 7 \times 41 \times 271\)

Combining Prime Factors in the Product

Now, let's write the prime factorization of the entire product by combining the factors we found:

Product \( = (7^1) \times (7^3) \times (5^1 \times 7^1 \times 11^1) \times (2^3 \times 5^3) \times (7^4) \times (7^1 \times 41^1 \times 271^1)\)

To find the total power of each prime factor in the product, we add the exponents for each base:

• Total power of 2: \(2^3\) (only from 1000)
• Total power of 5: \(5^1\) (from 385) + \(5^3\) (from 1000) \( = 5^{1+3} = 5^4\)
• Total power of 7: \(7^1\) (from 7) + \(7^3\) (from 343) + \(7^1\) (from 385) + \(7^4\) (from 2401) + \(7^1\) (from 77777) \( = 7^{1+3+1+4+1} = 7^{10}\)
• Total power of 11: \(11^1\) (from 385)
• Total power of 41: \(41^1\) (from 77777)
• Total power of 271: \(271^1\) (from 77777)

The prime factorization of the product is \(2^3 \times 5^4 \times 7^{10} \times 11^1 \times 41^1 \times 271^1\).

Determining the Maximum Value of n

For the product to be divisible by \(35^n = 5^n \times 7^n\), the number of factors of 5 in the product must be at least \(n\), and the number of factors of 7 in the product must be at least \(n\).

From the prime factorization of the product, we have:

• Number of factors of 5 is 4. So, we must have \(n \le 4\).
• Number of factors of 7 is 10. So, we must have \(n \le 10\).

For the product to be divisible by \(35^n\), both conditions must be satisfied simultaneously. Therefore, \(n\) must be less than or equal to the minimum of 4 and 10.

\(n \le \min(4, 10)\)

\(n \le 4\)

The maximum possible integer value for \(n\) that satisfies this condition is 4.

Step-by-Step Solution Summary

1. Identify the divisor: \(35^n\).
2. Find the prime factorization of the base of the divisor: \(35 = 5 \times 7\).
3. Rewrite the divisor using its prime factors: \(35^n = 5^n \times 7^n\).
4. Find the prime factorization of each number in the given product.
5. Combine the prime factors to find the total power of each prime in the product. Focus on the prime factors of the divisor (5 and 7).
6. Count the total number of factors of 5 in the product (which is 4).
7. Count the total number of factors of 7 in the product (which is 10).
8. For the product to be divisible by \(5^n \times 7^n\), \(n\) must be less than or equal to the power of 5 and less than or equal to the power of 7 in the product.
9. Determine the maximum \(n\) by finding the minimum of the powers of 5 and 7 (\(\min(4, 10) = 4\)).

The maximum value of \(n\) is 4.

Revision Table: Divisibility by Powers

Additional Information: Finding the Highest Power of a Prime Dividing a Factorial

A related concept in number theory involves finding the highest power of a prime number \(p\) that divides a factorial \(n!\). This is given by Legendre's formula:

\(E_p(n!) = \sum_{i=1}^{\infty} \lfloor \frac{n}{p^i} \rfloor = \lfloor \frac{n}{p} \rfloor + \lfloor \frac{n}{p^2} \rfloor + \lfloor \frac{n}{p^3} \rfloor + \dots\)

Where \(\lfloor x \rfloor\) is the floor function, giving the greatest integer less than or equal to \(x\).

While this specific formula wasn't directly used in solving the current problem, understanding how prime factors accumulate in products (like factorials) is a fundamental skill for these types of divisibility questions. The core idea remains the same: break down numbers into their prime factors and count the total occurrences of the relevant primes.