Question

What is the highest power of 12 in 99!?

Hint:

When finding the highest power of a composite number in a factorial, always find the powers of its prime factors. The answer is determined by the prime factor that is the "most scarce" relative to how many are needed for the composite number. For \(12 = 2^2 \times 3\), you compare the number of available 3s with half the number of available 2s.

Answer 1

Step 1: Understanding the Concept:
To find the highest power of a composite number (like 12) in a factorial (like 99!), we first need to find the prime factorization of the number. Then, we find the highest power of each of its prime factors in the factorial using Legendre's Formula. The limiting prime factor will determine the answer.
Step 2: Key Formula or Approach:
The highest power of a prime \(p\) in \(n!\), denoted \(E_p(n!)\), is given by Legendre's Formula: \[ E_p(n!) = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \dots \] Step 3: Detailed Explanation:
First, find the prime factorization of 12. \[ 12 = 2^2 \times 3^1 \] To form one factor of 12, we need two factors of 2 and one factor of 3.
Next, find the highest power of 2 in 99! using Legendre's Formula. \[ E_2(99!) = \left\lfloor \frac{99}{2} \right\rfloor + \left\lfloor \frac{99}{4} \right\rfloor + \left\lfloor \frac{99}{8} \right\rfloor + \left\lfloor \frac{99}{16} \right\rfloor + \left\lfloor \frac{99}{32} \right\rfloor + \left\lfloor \frac{99}{64} \right\rfloor \] \[ E_2(99!) = 49 + 24 + 12 + 6 + 3 + 1 = 95 \] So, the prime factorization of 99! contains \(2^{95}\).
Now, find the highest power of 3 in 99! using Legendre's Formula. \[ E_3(99!) = \left\lfloor \frac{99}{3} \right\rfloor + \left\lfloor \frac{99}{9} \right\rfloor + \left\lfloor \frac{99}{27} \right\rfloor + \left\lfloor \frac{99}{81} \right\rfloor \] \[ E_3(99!) = 33 + 11 + 3 + 1 = 48 \] So, the prime factorization of 99! contains \(3^{48}\).
Step 4: Final Answer:
We have 95 factors of 2 and 48 factors of 3 available. To form \(12 = 2^2 \times 3\), we need groups of \((2^2, 3)\). From the available \(3^{48}\), we can get 48 factors of 3. From the available \(2^{95}\), we can form \(\left\lfloor \frac{95}{2} \right\rfloor = 47\) factors of \(2^2\).
The number of factors of \(2^2\) (47) is less than the number of factors of 3 (48). Therefore, the number of pairs of 2s is the limiting factor. We can form a maximum of 47 groups of \((2^2, 3)\). Thus, the highest power of 12 in 99! is 47.