Understanding the Factorial Equation

The problem asks for the maximum possible integer value for \(k\) in the equation \(26! = n \cdot 8^k\), where \(n\) and \(k\) are positive integers. This requires finding the highest power of 8 that is a factor of \(26!\).

Calculating the Power of Prime Factor 2

To find the power of 8 (\(8 = 2^3\)), we first determine the total power of the prime factor 2 in the prime factorization of \(26!\). We can use Legendre's formula, which states that the exponent of a prime \(p\) in the prime factorization of \(N!\) is:

\(\text{Exponent of } p = \sum_{i=1}^{\infty} \left\lfloor \frac{N}{p^i} \right\rfloor\)

Here, \(N=26\) and \(p=2\). Let's calculate the terms:

• \(\lfloor \frac{26}{2} \rfloor = 13\)
• \(\lfloor \frac{26}{4} \rfloor = 6\)
• \(\lfloor \frac{26}{8} \rfloor = 3\)
• \(\lfloor \frac{26}{16} \rfloor = 1\)
• \(\lfloor \frac{26}{32} \rfloor = 0\) (and subsequent terms are 0)

Summing these values gives the exponent of 2 in \(26!\):

Total exponent of 2 = \(13 + 6 + 3 + 1 = 23\).

This means \(26!\) can be written as \(M \cdot 2^{23}\), where \(M\) is an integer not divisible by 2.

Relating Powers of 2 and 8

The given equation is \(26! = n \cdot 8^k\). Since \(8 = 2^3\), we can rewrite \(8^k\) as \((2^3)^k = 2^{3k}\).

So, the equation becomes \(26! = n \cdot 2^{3k}\).

We know \(26!\) contains the factor \(2^{23}\). For the equation to hold, the power of 2 on the right side, \(3k\) (assuming \(n\) is odd for maximum \(k\)), must be less than or equal to the power of 2 on the left side, 23.

\(3k \le 23\)

Finding the Maximum Integer Value of k

To find the maximum integer value of \(k\), we solve the inequality:

\(k \le \frac{23}{3}\)

\(k \le 7.666...\)

Since \(k\) must be a positive integer, the largest integer value \(k\) can take is 7.

Verification

If \(k=7\), then \(8^k = 8^7 = (2^3)^7 = 2^{21}\).

The equation is \(26! = n \cdot 2^{21}\).

Solving for \(n\), we get \(n = \frac{26!}{2^{21}}\).

Since \(26!\) contains \(2^{23}\) as a factor (\(26! = C \cdot 2^{23}\) for some integer \(C\)), we have:

\(n = \frac{C \cdot 2^{23}}{2^{21}} = C \cdot 2^{23-21} = C \cdot 2^2 = 4C\).

As \(26!\) is positive, \(C\) is a positive integer, making \(n=4C\) a positive integer. This confirms that \(k=7\) is a valid solution.

Therefore, the maximum value of \(k\) is 7.