The number `24!` is divisible by

Options

A. `6^(24)`
B. `24^(6)`
C. `12^(12)`
D. `48^(5)`

Answer

To determine the divisibility of \( 24! \) by various powers of numbers, we will follow these steps:

### Step 1: Calculate the highest power of 2 that divides \( 24! \)

To find the highest power of a prime \( p \) that divides \( n! \), we use the formula:

\[ \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor \]

For \( p = 2 \) and \( n = 24 \):

\[ \left\lfloor \frac{24}{2} \right\rfloor + \left\lfloor \frac{24}{4} \right\rfloor + \left\lfloor \frac{24}{8} \right\rfloor + \left\lfloor \frac{24}{16} \right\rfloor \]

Calculating each term:

- \( \left\lfloor \frac{24}{2} \right\rfloor = 12 \)
- \( \left\lfloor \frac{24}{4} \right\rfloor = 6 \)
- \( \left\lfloor \frac{24}{8} \right\rfloor = 3 \)
- \( \left\lfloor \frac{24}{16} \right\rfloor = 1 \)

Adding these values together gives:

\[ 12 + 6 + 3 + 1 = 22 \]

Thus, the highest power of 2 that divides \( 24! \) is \( 2^{22} \).

### Step 2: Calculate the highest power of 3 that divides \( 24! \)

Now we apply the same formula for \( p = 3 \):

\[ \left\lfloor \frac{24}{3} \right\rfloor + \left\lfloor \frac{24}{9} \right\rfloor + \left\lfloor \frac{24}{27} \right\rfloor \]

Calculating each term:

- \( \left\lfloor \frac{24}{3} \right\rfloor = 8 \)
- \( \left\lfloor \frac{24}{9} \right\rfloor = 2 \)
- \( \left\lfloor \frac{24}{27} \right\rfloor = 0 \)

Adding these values together gives:

\[ 8 + 2 + 0 = 10 \]

Thus, the highest power of 3 that divides \( 24! \) is \( 3^{10} \).

### Step 3: Combine the results

From the calculations, we have:

- \( 24! \) is divisible by \( 2^{22} \)
- \( 24! \) is divisible by \( 3^{10} \)

### Step 4: Determine the correct option

We need to check the options provided to see which one can be expressed in terms of \( 2^{22} \) and \( 3^{10} \).

1. **Option A**: \( 6^{10} = (2 \cdot 3)^{10} = 2^{10} \cdot 3^{10} \) (not sufficient in powers of 2)
2. **Option B**: \( 24^6 = (2^3 \cdot 3)^{6} = 2^{18} \cdot 3^{6} \) (not sufficient in powers of 2)
3. **Option C**: \( 12^{10} = (2^2 \cdot 3)^{10} = 2^{20} \cdot 3^{10} \) (not sufficient in powers of 2)
4. **Option D**: \( 48^5 = (2^4 \cdot 3)^{5} = 2^{20} \cdot 3^{5} \) (not sufficient in powers of 2)

The correct option based on our calculations is **Option B**, as it is the only one that satisfies the conditions of divisibility.

### Final Answer
The number \( 24! \) is divisible by \( 2^{22} \) and \( 3^{10} \), and the correct option is **Option B**.

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