Let the product of all the divisors of 1440 be P . If P is divisible by `24^(x)` , then the maximum value of x is :

Options

A. 28
B. 30
C. 32
D. 36

Answer

To solve the problem, we need to find the maximum value of \( x \) such that the product of all divisors of 1440, denoted as \( P \), is divisible by \( 24^x \).

### Step 1: Prime Factorization of 1440
First, we need to factor 1440 into its prime factors.

\[ 1440 = 144 \times 10 = (12^2) \times (2 \times 5) = (2^2 \times 3)^2 \times (2 \times 5) = 2^4 \times 3^2 \times 5^1 \]

Thus, the prime factorization of 1440 is:

\[ 1440 = 2^5 \times 3^2 \times 5^1 \]

### Step 2: Number of Divisors
Next, we calculate the number of divisors \( d(n) \) of a number \( n = p_1^{e_1} \times p_2^{e_2} \times \ldots \times p_k^{e_k} \) using the formula:

\[ d(n) = (e_1 + 1)(e_2 + 1) \ldots (e_k + 1) \]

For \( 1440 = 2^5 \times 3^2 \times 5^1 \):

\[ d(1440) = (5 + 1)(2 + 1)(1 + 1) = 6 \times 3 \times 2 = 36 \]

### Step 3: Product of Divisors
The product of all divisors \( P \) of a number \( n \) can be calculated using the formula:

\[ P = n^{d(n)/2} \]

Thus, for our case:

\[ P = 1440^{36/2} = 1440^{18} \]

### Step 4: Expressing \( P \) in terms of \( 24^x \)
Next, we need to express \( 1440^{18} \) in terms of \( 24^x \). First, we factor \( 24 \):

\[ 24 = 2^3 \times 3^1 \]

So, \( 24^x = (2^3 \times 3^1)^x = 2^{3x} \times 3^x \).

### Step 5: Finding the Powers of 2 and 3 in \( P \)
Now, we can express \( 1440^{18} \):

\[ 1440^{18} = (2^5 \times 3^2 \times 5^1)^{18} = 2^{90} \times 3^{36} \times 5^{18} \]

### Step 6: Setting Up the Inequality
For \( P \) to be divisible by \( 24^x \), we need:

\[ 2^{90} \times 3^{36} \text{ must be divisible by } 2^{3x} \times 3^x \]

This gives us two inequalities:

1. \( 90 \geq 3x \)
2. \( 36 \geq x \)

### Step 7: Solving the Inequalities
From the first inequality:

\[ 90 \geq 3x \implies x \leq 30 \]

From the second inequality:

\[ 36 \geq x \implies x \leq 36 \]

### Step 8: Conclusion
The maximum value of \( x \) that satisfies both inequalities is:

\[ x = 30 \]

Thus, the maximum value of \( x \) such that \( P \) is divisible by \( 24^x \) is:

\[ \boxed{30} \]