LCM of `12^(24),16^(18)` and N is `24^(24)` , then find the possible value of N? (a) 1825 (b) 320 (c) 563 (d) 468

Options

A. 1825
B. 320
C. 563
D. 468

Answer

To solve the problem of finding the possible value of \( N \) given that the LCM of \( 12^{24} \), \( 16^{18} \), and \( N \) is \( 24^{24} \), we will follow these steps:

### Step 1: Prime Factorization
First, we need to find the prime factorization of the numbers involved.

- \( 12 = 2^2 \times 3^1 \)
- \( 16 = 2^4 \)
- \( 24 = 2^3 \times 3^1 \)

### Step 2: Write the Powers
Now, we can express the powers of these numbers:

- \( 12^{24} = (2^2 \times 3^1)^{24} = 2^{48} \times 3^{24} \)
- \( 16^{18} = (2^4)^{18} = 2^{72} \)
- \( N = 2^a \times 3^b \) (where \( a \) and \( b \) are the powers of 2 and 3 in \( N \))

### Step 3: Find the LCM
The LCM of multiple numbers is found by taking the highest power of each prime factor present in any of the numbers.

The LCM is given as \( 24^{24} = (2^3 \times 3^1)^{24} = 2^{72} \times 3^{24} \).

### Step 4: Set Up the Equations
From the LCM, we can set up the following equations based on the prime factorization:

1. For the prime factor \( 2 \):
\[ \max(48, 72, a) = 72 \]
This means \( a \) can be at most \( 72 \) (it can be less than or equal to \( 72 \)).

2. For the prime factor \( 3 \):
\[ \max(24, b) = 24 \]
This means \( b \) can be at most \( 24 \) (it can be less than or equal to \( 24 \)).

### Step 5: Determine Possible Values of \( N \)
Since \( N \) can take values based on \( a \) and \( b \), we can express \( N \) as:
\[ N = 2^a \times 3^b \]
where \( 0 \leq a \leq 72 \) and \( 0 \leq b \leq 24 \).

### Step 6: Evaluate Options
Now we need to check the options provided to see which one can be expressed in the form \( 2^a \times 3^b \):

(a) 1825 = \( 5^2 \times 73 \) (not valid)

(b) 320 = \( 2^6 \times 5^1 \) (not valid)

(c) 563 = \( 563^1 \) (not valid)

(d) 468 = \( 2^2 \times 3^2 \times 13^1 \) (not valid)

None of the options seem to fit the required form of \( N = 2^a \times 3^b \).

### Conclusion
Upon reviewing the calculations and the options, it seems that the problem may have a misalignment with the options provided. However, based on the calculations, we can conclude that \( N \) must be of the form \( 2^a \times 3^b \) where \( a \) can be up to \( 72 \) and \( b \) can be up to \( 24 \).