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-by-Step Solution:

1. **Prime Factorization of 1440**:
We start by factorizing \( 1440 \) into its prime factors.
\[ 1440 = 144 \times 10 = (12^2) \times (2 \times 5) = (2^2 \times 3)^2 \times (2 \times 5) = 2^5 \times 3^2 \times 5^1 \]
Thus, the prime factorization of \( 1440 \) is:
\[ 1440 = 2^5 \times 3^2 \times 5^1 \]

2. **Finding the Number of Divisors**:
The number of divisors \( d(n) \) of a number \( n = p_1^{e_1} \times p_2^{e_2} \times ... \times p_k^{e_k} \) is given by:
\[ d(n) = (e_1 + 1)(e_2 + 1)...(e_k + 1) \]
For \( 1440 = 2^5 \times 3^2 \times 5^1 \):
\[ d(1440) = (5 + 1)(2 + 1)(1 + 1) = 6 \times 3 \times 2 = 36 \]

3. **Product of All Divisors**:
The product of all divisors of \( n \) can be calculated using the formula:
\[ P = n^{d(n)/2} \]
Therefore, for \( 1440 \):
\[ P = 1440^{36/2} = 1440^{18} \]

4. **Expressing \( P \) in Terms of Prime Factors**:
We can express \( P \) as:
\[ P = (2^5 \times 3^2 \times 5^1)^{18} = 2^{90} \times 3^{36} \times 5^{18} \]

5. **Finding the Maximum \( x \) for \( 24^x \)**:
We know that:
\[ 24 = 2^3 \times 3^1 \]
Therefore:
\[ 24^x = (2^3 \times 3^1)^x = 2^{3x} \times 3^x \]
For \( P \) to be divisible by \( 24^x \), we need:
\[ 2^{90} \text{ must be greater than or equal to } 2^{3x} \quad \text{and} \quad 3^{36} \text{ must be greater than or equal to } 3^x \]
This gives us two inequalities:
\[ 90 \geq 3x \quad \Rightarrow \quad x \leq 30 \]
\[ 36 \geq x \quad \Rightarrow \quad x \leq 36 \]

6. **Finding the Maximum \( x \)**:
The maximum value of \( x \) that satisfies both inequalities is:
\[ x \leq 30 \]

Thus, the maximum value of \( x \) is \( \boxed{30} \).